MathsCBSE Class 10 · Chapter 6

Triangles Class 10: Similar Triangles, BPT and Similarity Criteria

Updated 13 October 2026 · Fact-checked

Triangles in Class 10 is about similar triangles: triangles with the same shape, equal angles and proportional sides. You must prove the Basic Proportionality Theorem and use its converse and the AA, SSS and SAS criteria. The board asks you to find unknown lengths, prove two triangles similar, and solve real-life problems such as shadows and heights.

What Triangles is about

This chapter is about similarity: when two figures have the same shape but possibly different sizes. You learn what similar figures are, how to recognise similar triangles, and the rules (criteria) that let you prove two triangles similar by checking only a few facts. The one theorem you must prove is the Basic Proportionality Theorem (BPT), also called Thales theorem. Its converse and the AA, SSS and SAS criteria are stated without proof and used freely.

The chapter builds on congruence of triangles from Class 9, and on parallel lines and angles (alternate and corresponding angles). It feeds directly into Coordinate Geometry (section formula ideas), Circles (proofs with tangents and chords often use similar triangles) and Some Applications of Trigonometry, where heights and distances work because right triangles with equal angles are similar.

Board questions take a few shapes: find an unknown length or a value of $x$ using BPT; decide whether a line is parallel to a side; prove two triangles similar and deduce a ratio or product of sides; write the correct correspondence from given sides; assertion-reason items on the criteria; the full proof of BPT as a long answer; and case-based questions on shadows, lamp posts, ramps and scale drawings.

Weightage: This chapter is part of Unit IV, Geometry, which carries 15 of the 80 theory marks together with Circles.Source: CBSE Mathematics syllabus 2026-27, Unit IV Geometry

In real board papers: Standard, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 8.2 marks per paper; Basic, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 8.4 marks per paper. See the questions

Key concepts

Similar figures: definition, examples and counter examples

NCERT section 6.2 Similar Figures

Two figures are similar if they have the same shape but not necessarily the same size. For two polygons with the same number of sides, this needs two conditions together: (i) corresponding angles are equal, and (ii) corresponding sides are in the same ratio.

Both conditions matter, and the board tests this with counter examples:

A square of side 5 cm and a rhombus of side 5 cm have sides in ratio 1 : 1, but their angles differ, so they are not similar.
A 6 cm by 4 cm rectangle and a 6 cm by 6 cm square have every angle $9 0^{\circ}$ , but their sides are not in one ratio, so they are not similar.

All circles, all squares and all equilateral triangles are similar, because a single measurement (radius or side) fixes the shape. Congruent figures are a special case of similar figures where the ratio of sides is 1.

Worked example

Which of the following pairs of figures is always similar?
(a) Two rhombuses
(b) Two rectangles
(c) Two squares
(d) Two isosceles triangles

Try it first, then show the solutionSolution

Rhombuses: all sides equal, so sides are proportional, but angles can differ (one can have $6 0^{\circ}$ , another $9 0^{\circ}$ ). Not always similar.
Rectangles: all angles are $9 0^{\circ}$ , but a 4 by 2 rectangle and a 4 by 3 rectangle have ratios $\frac{4}{4} \neq = \frac{2}{3}$ . Not always similar.
Isosceles triangles: one can have angles $4 0^{\circ}, 7 0^{\circ}, 7 0^{\circ}$ , another $10 0^{\circ}, 4 0^{\circ}, 4 0^{\circ}$ . Not always similar.
Squares: every angle is $9 0^{\circ}$ and all four sides of one square divided by those of the other give the same ratio. Both conditions always hold.

Answer: (c) Two squares are always similar.

Similar triangles and correspondence

NCERT section 6.3 Similarity of Triangles

$△ A B C \sim △ P QR$ means $∠ A = ∠ P$ , $∠ B = ∠ Q$ , $∠ C = ∠ R$ and $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{C A}{R P}$ .

The order of the letters is the correspondence: first letter matches first, second matches second, third matches third. To find which side matches $A B$ , take the first two letters of the other triangle: $P Q$ . Writing the letters in the wrong order gives wrong ratios and loses marks.

The common ratio is the scale factor $k$ . Since every side is multiplied by the same $k$ , the perimeter is also multiplied by $k$ . So the ratio of perimeters equals the ratio of corresponding sides. The same holds for corresponding altitudes and medians.

For triangles, equal angles and proportional sides always come together, which is why the criteria later need you to check only some of the facts.

Worked example

$△ A B C \sim △ D E F$ with $A B = 4$ cm, $B C = 6$ cm and $C A = 5$ cm. The perimeter of $△ D E F$ is 45 cm. Find $D E$ , $E F$ and $F D$ .

Try it first, then show the solutionSolution

Perimeter of $△ A B C = 4 + 6 + 5 = 15$ cm.
Ratio of perimeters = ratio of corresponding sides, so $\frac{D E}{A B} = \frac{E F}{B C} = \frac{F D}{C A} = \frac{45}{15} = 3$ .
$D E = 3 \times 4 = 12$ cm (DE corresponds to AB).
$E F = 3 \times 6 = 18$ cm and $F D = 3 \times 5 = 15$ cm.
Check: $12 + 18 + 15 = 45$ cm.

Answer: $D E = 12$ cm, $E F = 18$ cm, $F D = 15$ cm.

Basic Proportionality Theorem (Thales theorem) with proof

NCERT section 6.3 Similarity of Triangles

Statement: If a line is drawn parallel to one side of a triangle to meet the other two sides in distinct points, it divides those two sides in the same ratio. In $△ A B C$ with $D E ∥ B C$ , D on AB and E on AC: $\frac{A D}{D B} = \frac{A E}{E C}$ .

Proof idea: Join BE and CD. Draw $E N ⊥ A B$ and $D M ⊥ A C$ .

$\frac{a r ( A D E )}{a r ( B D E )} = \frac{\frac{1}{2} \cdot A D \cdot E N}{\frac{1}{2} \cdot D B \cdot E N} = \frac{A D}{D B}$ , as both triangles have the same height EN.
Similarly $\frac{a r ( A D E )}{a r ( D E C )} = \frac{A E}{E C}$ , with common height DM.
$a r (B D E) = a r (D E C)$ , because they stand on the same base DE and lie between the same parallels DE and BC.

So the two ratios are equal. Use BPT whenever a line inside a triangle is parallel to a side and you need a missing length on the other two sides.

Worked example

In $△ A B C$ , D and E are points on AB and AC such that $D E ∥ B C$ . If $A D = x$ cm, $D B = (x - 2)$ cm, $A E = (x + 2)$ cm and $E C = (x - 1)$ cm, find $x$ and the lengths AB and AC.

Try it first, then show the solutionSolution

Since $D E ∥ B C$ , by BPT $\frac{A D}{D B} = \frac{A E}{E C}$ .
$\frac{x}{x - 2} = \frac{x + 2}{x - 1}$
Cross-multiplying: $x (x - 1) = (x - 2) (x + 2)$
$x^{2} - x = x^{2} - 4$ , so $x = 4$ .
$A D = 4$ , $D B = 2$ , $A E = 6$ , $E C = 3$ ; check $\frac{4}{2} = \frac{6}{3} = 2$ .
$A B = 4 + 2 = 6$ cm and $A C = 6 + 3 = 9$ cm.

Answer: $x = 4$ , $A B = 6$ cm, $A C = 9$ cm.

Converse of the Basic Proportionality Theorem

NCERT section 6.3 Similarity of Triangles

Statement (no proof needed): If a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side. In $△ A B C$ , if $\frac{A D}{D B} = \frac{A E}{E C}$ , then $D E ∥ B C$ .

BPT goes from parallel to ratio. The converse goes from ratio to parallel. Use it when a question asks you to show a line is parallel, or when parallel lines are not given but lengths are.

Always compare like with like: part to part ( $\frac{A D}{D B}$ with $\frac{A E}{E C}$ ) or part to whole ( $\frac{A D}{A B}$ with $\frac{A E}{A C}$ ). Comparing a part on one side with a whole on the other is a classic mistake.

Once you know $D E ∥ B C$ , corresponding angles are equal, so $△ A D E \sim △ A B C$ by AA. That gives $\frac{D E}{B C} = \frac{A D}{A B}$ , which BPT alone cannot give you.

Worked example

In $△ P QR$ , S is a point on PQ and T is a point on PR. $P Q = 10$ cm, $P S = 4$ cm, $P R = 15$ cm and $T R = 9$ cm. Show that $S T ∥ QR$ . If $QR = 20$ cm, find ST.

Try it first, then show the solutionSolution

$S Q = P Q - P S = 10 - 4 = 6$ cm and $P T = P R - T R = 15 - 9 = 6$ cm.
$\frac{P S}{S Q} = \frac{4}{6} = \frac{2}{3}$ and $\frac{P T}{T R} = \frac{6}{9} = \frac{2}{3}$ .
Since $\frac{P S}{S Q} = \frac{P T}{T R}$ , $S T ∥ QR$ (converse of BPT).
In $△ P S T$ and $△ P QR$ : $∠ P$ is common and $∠ P S T = ∠ P QR$ (corresponding angles, $S T ∥ QR$ ).
So $△ P S T \sim △ P QR$ (AA similarity).
$\frac{S T}{QR} = \frac{P S}{P Q} = \frac{4}{10}$ , so $S T = \frac{4}{10} \times 20 = 8$ cm.

Answer: $S T ∥ QR$ and $S T = 8$ cm.

AAA and AA similarity criterion

NCERT section 6.4 Criteria for Similarity of Triangles

Statement (no proof needed): If in two triangles the corresponding angles are equal, then their corresponding sides are proportional and the triangles are similar (AAA).

You only need two angles. The angle sum of a triangle is $18 0^{\circ}$ , so if two angles match, the third must match too. This shortened version is called AA similarity, and it is the criterion you will use most often.

Where do equal angles come from? Look for:

a common angle shared by both triangles
vertically opposite angles
alternate or corresponding angles from parallel lines
right angles from perpendiculars

A common board setup is a trapezium: the diagonals cut each other into two similar triangles, so the diagonals are divided in the ratio of the parallel sides.

Worked example

ABCD is a trapezium with $A B ∥ D C$ . Its diagonals AC and BD meet at O. $A B = 12$ cm, $D C = 8$ cm, $A O = 6$ cm and $B D = 15$ cm. Find OC, BO and OD.

Try it first, then show the solutionSolution

In $△ A O B$ and $△ C O D$ : $∠ O A B = ∠ O C D$ (alternate angles, $A B ∥ D C$ , transversal AC).
$∠ A O B = ∠ C O D$ (vertically opposite angles).
So $△ A O B \sim △ C O D$ (AA similarity).
Hence $\frac{A O}{C O} = \frac{B O}{D O} = \frac{A B}{C D} = \frac{12}{8} = \frac{3}{2}$ .
$\frac{6}{C O} = \frac{3}{2}$ , so $C O = 4$ cm.
$B O : O D = 3 : 2$ and $B O + O D = 15$ , so $B O = \frac{3}{5} \times 15 = 9$ cm and $O D = 6$ cm.

Answer: $O C = 4$ cm, $B O = 9$ cm, $O D = 6$ cm.

SSS similarity criterion

NCERT section 6.4 Criteria for Similarity of Triangles

Statement (no proof needed): If the corresponding sides of two triangles are proportional, then their corresponding angles are equal and the triangles are similar.

Use SSS when you are given all six side lengths and no angles. The hard part is the correspondence when sides are listed in a jumbled order. The safe method:

Sort each triangle's sides from smallest to largest.
Divide smallest by smallest, middle by middle, largest by largest.
If all three ratios are equal, the triangles are similar.

Then fix the vertex order. A vertex is where two sides meet, so the vertex shared by two sides in one triangle matches the vertex shared by the matching two sides in the other. The angle opposite the smallest side matches the angle opposite the smallest side, and so on. Write the statement in that order before using any angle.

Worked example

In $△ A B C$ , $A B = 6$ cm, $B C = 9$ cm, $C A = 12$ cm. In $△ D E F$ , $D E = 16$ cm, $E F = 8$ cm, $F D = 12$ cm. Show that the triangles are similar, write the similarity statement in the correct order, name the angle of $△ A B C$ equal to $∠ D$ , and find the ratio of their perimeters.

Try it first, then show the solutionSolution

Sorted: $△ A B C$ : 6, 9, 12 ( $A B, B C, C A$ ). $△ D E F$ : 8, 12, 16 ( $E F, F D, D E$ ).
$\frac{A B}{E F} = \frac{6}{8} = \frac{3}{4}$ , $\frac{B C}{F D} = \frac{9}{12} = \frac{3}{4}$ , $\frac{C A}{D E} = \frac{12}{16} = \frac{3}{4}$ .
All ratios are equal, so the triangles are similar (SSS similarity).
B is common to AB and BC; F is common to EF and FD. So $B \leftrightarrow F$ . Then $A \leftrightarrow E$ and $C \leftrightarrow D$ .
Statement: $△ A B C \sim △ E F D$ , so $∠ C = ∠ D$ .
Perimeters: $6 + 9 + 12 = 27$ cm and $16 + 8 + 12 = 36$ cm; ratio $27 : 36 = 3 : 4$ , matching the side ratio.

Answer: $△ A B C \sim △ E F D$ ; $∠ C = ∠ D$ ; ratio of perimeters $= 3 : 4$ .

SAS similarity criterion

NCERT section 6.4 Criteria for Similarity of Triangles

Statement (no proof needed): If one angle of a triangle equals one angle of another triangle and the sides including these angles are proportional, the triangles are similar.

"Including" means the angle lies between the two sides. If $∠ A = ∠ P$ , you must compare $\frac{A B}{P Q}$ and $\frac{A C}{P R}$ , the two sides that form angle A and angle P. If the equal angle is not between the proportional sides (an SSA situation), the triangles need not be similar, and the board penalises this.

SAS is the tool when two triangles share an angle and you know two sides of each, for example a small triangle cut off at a vertex. Watch out for a reversed correspondence: sometimes the small triangle matches the big one with its sides swapped, as in the example below, so $\frac{A D}{A C}$ , not $\frac{A D}{A B}$ .

Worked example

In $△ A B C$ , $A B = 8$ cm, $A C = 6$ cm and $B C = 10$ cm. D is a point on AB with $A D = 3$ cm and E is a point on AC with $A E = 4$ cm. Prove that $△ A D E \sim △ A C B$ and find DE.

Try it first, then show the solutionSolution

$\frac{A D}{A C} = \frac{3}{6} = \frac{1}{2}$ and $\frac{A E}{A B} = \frac{4}{8} = \frac{1}{2}$ .
So $\frac{A D}{A C} = \frac{A E}{A B}$ .
$∠ D A E = ∠ C A B$ (common angle A, included between these sides).
Hence $△ A D E \sim △ A C B$ (SAS similarity).
Corresponding sides: $\frac{D E}{C B} = \frac{A D}{A C} = \frac{1}{2}$ .
$D E = \frac{1}{2} \times 10 = 5$ cm.

Answer: $△ A D E \sim △ A C B$ and $D E = 5$ cm.

Applying similarity to real-life problems

NCERT section 6.4 Criteria for Similarity of Triangles

Many case-based questions hide two similar triangles inside a real situation. At one moment, the sun's rays make the same angle with the ground for every object, and objects stand vertically, so object-and-shadow triangles are similar by AA. A lamp post and a person standing on level ground also give similar triangles: both are vertical (angle $9 0^{\circ}$ ), and they share the angle at the tip of the shadow.

The method:

Draw the situation and label every point.
Name the two right angles and the common or equal angle; state AA.
Write the ratio with the correct whole lengths. With a lamp post, the post's triangle has base distance + shadow, not just the distance.
Solve and give units.

The same idea works for mirrors (angle of incidence equals angle of reflection), ramps and scale models.

Worked example

A boy 1.6 m tall stands 6 m from the foot of a vertical lamp post 4.8 m high on level ground. Find the length of his shadow. How much further from the post must he walk for his shadow to become 4 m long?

Try it first, then show the solutionSolution

Let the post be AB (B on ground), the boy CD (D on ground), and E the tip of the shadow, with B, D, E in a line. $B D = 6$ m, $D E = s$ .
In $△ E D C$ and $△ E B A$ : $∠ E D C = ∠ E B A = 9 0^{\circ}$ and $∠ E$ is common.
So $△ E D C \sim △ E B A$ (AA similarity).
$\frac{C D}{A B} = \frac{E D}{E B}$ , so $\frac{1.6}{4.8} = \frac{s}{s + 6}$ .
$\frac{1}{3} = \frac{s}{s + 6}$ , so $s + 6 = 3 s$ and $s = 3$ m.
For a 4 m shadow at distance $d$ : $\frac{1}{3} = \frac{4}{4 + d}$ , so $4 + d = 12$ and $d = 8$ m.
He must move $8 - 6 = 2$ m further away. Check: $\frac{4}{12} = \frac{1.6}{4.8}$ .

Answer: Shadow = 3 m; he must walk 2 m further away from the post.

Maths in real life: Triangles

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Basic Proportionality Theorem (Thales theorem) with proof

Imagine your family draws a large triangular rangoli ABC for Diwali. A coloured stripe DE runs parallel to the base BC, with D on AB and E on AC. Your mother's measurements, in cm, are AD = x, DB = x − 15, AE = x + 15 and EC = x − 5. The base BC is 90 cm long.

(i) Find $x$ and the lengths of sides AB and AC. (ii) Find the length of the stripe DE. (iii) You want to outline the small triangle ADE with ribbon. How much ribbon do you need?

Try it first, then show the solutionSolution

$D E ∥ B C$ , so by the Basic Proportionality Theorem, $\frac{A D}{D B} = \frac{A E}{E C}$ .
$\frac{x}{x - 15} = \frac{x + 15}{x - 5}$
Cross-multiplying: $x (x - 5) = (x - 15) (x + 15)$ , so $x^{2} - 5 x = x^{2} - 225$ .
$5 x = 225$ , so $x = 45$ .
$A D = 45$ cm, $D B = 30$ cm, $A E = 60$ cm, $E C = 40$ cm. Check: $\frac{45}{30} = \frac{60}{40} = \frac{3}{2}$ .
$A B = 45 + 30 = 75$ cm and $A C = 60 + 40 = 100$ cm.
(ii) $∠ A$ is common and $∠ A D E = ∠ A B C$ (corresponding angles, $D E ∥ B C$ ), so $△ A D E \sim △ A B C$ (AA).
$\frac{D E}{B C} = \frac{A D}{A B} = \frac{45}{75} = \frac{3}{5}$ , so $D E = \frac{3}{5} \times 90 = 54$ cm.
(iii) Ribbon $= A D + A E + D E = 45 + 60 + 54 = 159$ cm.

Answer: $x = 45$ ; AB = 75 cm and AC = 100 cm; the stripe DE is 54 cm long; you need 159 cm of ribbon.

Memory hook: A rangoli stripe parallel to the base splits both sides in the same ratio; for the stripe's own length, use the whole side.

Converse of the Basic Proportionality Theorem

Suppose an A-frame step ladder stands on a level floor with its top at A and its feet at B and C, which are 100 cm apart. A tie-bar DE joins point D on leg AB to point E on leg AC. The measurements are AD = 90 cm, DB = 60 cm, AE = 105 cm and EC = 70 cm.

(i) Show that the tie-bar is parallel to the floor BC. (ii) Find the length of the tie-bar. (iii) A second bar, 80 cm long and also parallel to the floor, is to be fitted lower down. How far from the top A must it be fixed along each leg?

Try it first, then show the solutionSolution

(i) $\frac{A D}{D B} = \frac{90}{60} = \frac{3}{2}$ and $\frac{A E}{E C} = \frac{105}{70} = \frac{3}{2}$ .
$\frac{A D}{D B} = \frac{A E}{E C}$ , so by the converse of BPT, $D E ∥ B C$ . The bar is parallel to the floor.
(ii) $A B = 90 + 60 = 150$ cm and $A C = 105 + 70 = 175$ cm.
$∠ A$ is common and $∠ A D E = ∠ A B C$ (corresponding angles, $D E ∥ B C$ ), so $△ A D E \sim △ A B C$ (AA).
$\frac{D E}{B C} = \frac{A D}{A B} = \frac{90}{150} = \frac{3}{5}$ , so $D E = \frac{3}{5} \times 100 = 60$ cm.
(iii) Let the new bar be FG with F on AB and G on AC. As $F G ∥ B C$ , $△ A F G \sim △ A B C$ (AA), so $\frac{F G}{B C} = \frac{A F}{A B} = \frac{A G}{A C}$ .
$\frac{80}{100} = \frac{A F}{150}$ , so $A F = 120$ cm.
$\frac{80}{100} = \frac{A G}{175}$ , so $A G = 140$ cm.
Check: $\frac{A F}{F B} = \frac{120}{30} = 4$ and $\frac{A G}{GC} = \frac{140}{35} = 4$ , equal, so the bar is indeed parallel.

Answer: (i) Both ratios equal $\frac{3}{2}$ , so $D E ∥ B C$ . (ii) The tie-bar is 60 cm long. (iii) Fix it 120 cm from A along AB and 140 cm from A along AC.

Memory hook: Same ratio on both legs means the bar is level (converse of BPT); then AA gives its length.

AAA and AA similarity criterion

Suppose your school builds a straight wheelchair ramp. It starts at point A on level ground and rises to the doorway step C, which is 1 m high and directly above point B on the ground. The builders have not yet measured AB. They know that a vertical support DE stands at D on AB with AD = 3 m, and its height is 0.25 m, its top E touching the ramp.

(i) Find the horizontal distance AB. (ii) A second vertical support must stand on AB exactly midway between D and B. Find its height. (iii) A third support is to be 0.8 m tall. How far beyond the second support should it stand?

Try it first, then show the solutionSolution

In $△ A D E$ and $△ A B C$ : $∠ A$ is common and $∠ A D E = ∠ A B C = 9 0^{\circ}$ (both supports are vertical).
So $△ A D E \sim △ A B C$ (AA), giving $\frac{D E}{B C} = \frac{A D}{A B}$ .
(i) $\frac{0.25}{1} = \frac{3}{A B}$ , so $A B = \frac{3}{0.25} = 12$ m.
(ii) $D B = 12 - 3 = 9$ m, so the midpoint F of DB is at $A F = 3 + 4.5 = 7.5$ m.
For the support FG, $△ A F G \sim △ A B C$ for the same reason, so $\frac{F G}{1} = \frac{7.5}{12}$ and $F G = 0.625$ m.
(iii) For a support of height 0.8 m at H: $\frac{0.8}{1} = \frac{A H}{12}$ , so $A H = 9.6$ m.
Distance beyond the second support $= 9.6 - 7.5 = 2.1$ m.
Check: $\frac{0.25}{3} = \frac{0.625}{7.5} = \frac{0.8}{9.6} = \frac{1}{12}$ .

Answer: (i) AB = 12 m. (ii) The second support is 0.625 m (62.5 cm) tall. (iii) It should stand 2.1 m beyond the second support, that is 9.6 m from A.

Memory hook: Every support under a ramp makes a mini copy of the big triangle: common angle plus right angle gives AA.

SSS similarity criterion

Imagine your class makes triangular cardboard signboards for a science fair. The small board has sides 30 cm, 40 cm and 50 cm, and the angle between the 30 cm and 40 cm sides is $9 0^{\circ}$ . A large board must have the same shape, and the teacher only fixes its perimeter at 180 cm. A third board, also the same shape, has lost its labels: you only know two of its sides are 54 cm and 90 cm.

(i) Find the three sides of the large board. (ii) Find the altitude of the large board drawn to its longest side. (iii) Find the third side of the third board and say which of its angles is $9 0^{\circ}$ .

Try it first, then show the solutionSolution

(i) Perimeter of the small board $= 30 + 40 + 50 = 120$ cm.
Perimeters of similar triangles are in the ratio of their sides, so the scale factor is $k = \frac{180}{120} = \frac{3}{2}$ .
Large board sides: $\frac{3}{2} \times 30 = 45$ cm, $\frac{3}{2} \times 40 = 60$ cm, $\frac{3}{2} \times 50 = 75$ cm. Check: $45 + 60 + 75 = 180$ cm.
(ii) Small board: area $= \frac{1}{2} \times 30 \times 40 = 600$ cm $^{2}$ , and also $\frac{1}{2} \times 50 \times h$ , so $h = 24$ cm.
Corresponding altitudes scale by $k$ , so the large board's altitude $= \frac{3}{2} \times 24 = 36$ cm. Check: $\frac{1}{2} \times 45 \times 60 = 1350 = \frac{1}{2} \times 75 \times 36$ .
(iii) The sides 54 and 90 must be $k$ times two sides of the small board with the same $k$ , so those sides are in the ratio $\frac{90}{54} = \frac{5}{3}$ .
Among 30, 40, 50 only $\frac{50}{30} = \frac{5}{3}$ (since $\frac{40}{30} = \frac{4}{3}$ and $\frac{50}{40} = \frac{5}{4}$ ). So $54 \leftrightarrow 30$ , $90 \leftrightarrow 50$ and $k = \frac{54}{30} = 1.8$ .
Third side $= 1.8 \times 40 = 72$ cm. All three ratios equal 1.8, so the boards are similar by SSS.
The 30 cm and 40 cm sides correspond to the 54 cm and 72 cm sides, so the angle between the 54 cm and 72 cm sides is $9 0^{\circ}$ .

Answer: (i) 45 cm, 60 cm and 75 cm. (ii) 36 cm. (iii) The third side is 72 cm, and the right angle is between the 54 cm and 72 cm sides.

Memory hook: Sort the sides and divide smallest by smallest: one common ratio means SSS, and every length, perimeter and altitude scales by it.

Applying similarity to real-life problems

Suppose a street lamp in your colony is 4.5 m tall. Your sister, who is 1.5 m tall, starts at the foot of the lamp post and walks away from it in a straight line on level ground. She walks at 1.2 m per second.

How long is her shadow after she has walked for 4 seconds? Show the similar triangles you use.

Try it first, then show the solutionSolution

Distance walked $= 1.2 \times 4 = 4.8$ m.
Let the lamp top be L, the lamp foot O, her head H, her feet F and the shadow tip S. Let the shadow FS $= s$ m.
In $△ S F H$ and $△ S O L$ : $∠ S$ is common.
$∠ S F H = ∠ S O L = 9 0^{\circ}$ , because she and the lamp post are both vertical.
So $△ S F H \sim △ S O L$ by AA similarity.
$\frac{F H}{O L} = \frac{S F}{S O}$ gives $\frac{1.5}{4.5} = \frac{s}{s + 4.8}$ .
$\frac{1}{3} = \frac{s}{s + 4.8}$ , so $s + 4.8 = 3 s$ .
$2 s = 4.8$ , so $s = 2.4$ m.
Check: $\frac{2.4}{2.4 + 4.8} = \frac{2.4}{7.2} = \frac{1}{3}$ . Correct.

Answer: Her shadow is 2.4 m long.

Memory hook: Under a lamp post, the big triangle's base is distance plus shadow, not just the shadow.

Topics in this chapter

Triangles formula sheet

Definition of similar triangles: $△ A B C \sim △ P QR ⟺ ∠ A = ∠ P, ∠ B = ∠ Q, ∠ C = ∠ R and \frac{A B}{P Q} = \frac{B C}{QR} = \frac{C A}{R P}$; The order of letters fixes which angles and sides match. Use it to read off ratios once similarity is proved.
Basic Proportionality Theorem (BPT): $D E ∥ B C \Rightarrow \frac{A D}{D B} = \frac{A E}{E C}$; D on AB, E on AC. Use when a line parallel to one side cuts the other two sides. This is the only theorem you must prove.
BPT, part-to-whole forms: $\frac{A D}{A B} = \frac{A E}{A C}, \frac{D B}{A B} = \frac{E C}{A C}$; Follow from BPT by adding 1 to each side of the reciprocal ratio. Useful when a whole side is given.
Converse of BPT: $\frac{A D}{D B} = \frac{A E}{E C} \Rightarrow D E ∥ B C$; Stated without proof. Use to show a line is parallel to the third side.
Third side of the small triangle: $D E ∥ B C \Rightarrow △ A D E \sim △ A B C \Rightarrow \frac{D E}{B C} = \frac{A D}{A B} = \frac{A E}{A C}$; Comes from AA similarity, not BPT. Use the whole side AB, never DB, in the ratio for DE.
AA (AAA) similarity criterion: $∠ A = ∠ P, ∠ B = ∠ Q \Rightarrow △ A B C \sim △ P QR$; Two equal angles are enough since the third follows from the angle sum $18 0^{\circ}$ . Most used criterion.
SSS similarity criterion: $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{C A}{R P} \Rightarrow △ A B C \sim △ P QR$; Use when all six sides are known. Match smallest with smallest, largest with largest.
SAS similarity criterion: $∠ A = ∠ P and \frac{A B}{P Q} = \frac{A C}{P R} \Rightarrow △ A B C \sim △ P QR$; The equal angle must be included between the two proportional sides. SSA is not a criterion.
Perimeters, altitudes and medians of similar triangles: $\frac{perimeter of △ A B C}{perimeter of △ P QR} = \frac{A B}{P Q} = \frac{altitude _{1}}{altitude _{2}} = \frac{median _{1}}{median _{2}}$; All corresponding lengths scale by the same factor. Do not extend this to areas; that result is not in the syllabus.
Diagonals of a trapezium: $A B ∥ D C, A C \cap B D = O \Rightarrow \frac{A O}{O C} = \frac{B O}{O D} = \frac{A B}{C D}$; From $△ A O B \sim △ C O D$ (AA). Write the similarity proof before using it.
Line parallel to the parallel sides of a trapezium: $A B ∥ E F ∥ D C, E on A D, F on B C \Rightarrow \frac{A E}{E D} = \frac{B F}{F C}$; Prove by joining a diagonal and applying BPT in each of the two triangles formed.
Shadow and lamp-post ratios: $\frac{h _{1}}{h _{2}} = \frac{s _{1}}{s _{2}} (sun), \frac{h}{H} = \frac{s}{s + d} (lamp post)$; $h$ = person's height, $H$ = lamp height, $s$ = shadow, $d$ = distance from post. Justify with AA first.
Angle sum of a triangle: $∠ A + ∠ B + ∠ C = 18 0^{\circ}$; Lets you find the third equal angle and explains why AA is enough.

How the board asks Triangles

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Find an unknown length or x using BPT	A line DE parallel to BC cuts AB and AC; parts of the sides are given as numbers or expressions in x. Find x or a missing length.	Write "Since $D E ∥ B C$ , by BPT" and then $\frac{A D}{D B} = \frac{A E}{E C}$ . Cross-multiply, solve, and check that every length comes out positive. If a whole side is given, use the part-to-whole form.
Show that a line is parallel (converse of BPT)	Lengths on two sides are given with no parallel lines stated. Show $D E ∥ B C$ , often followed by finding DE.	Compute the missing parts, find both part-to-part ratios, show they are equal, then quote the converse of BPT. For DE, prove $△ A D E \sim △ A B C$ by AA and use $\frac{D E}{B C} = \frac{A D}{A B}$ .
Prove two triangles similar and deduce a relation	"Prove that $△ \dots \sim △ \dots$ and hence show that $P A \cdot P B = P C \cdot P D$ " or similar product or square relations.	Pick the two triangles containing all four lengths in the result. List two pairs of equal angles with reasons (common, vertically opposite, alternate, $9 0^{\circ}$ ), state AA, write the ratio of sides in correct order, then cross-multiply.
Correspondence and similarity from given sides or angles	MCQs give two triangles with jumbled sides or angles and ask which statement is correct, or which angle equals a given one.	Sort sides by size, check the three ratios, then match vertices by the sides that meet there. Write the full similarity statement before choosing an option.
Trapezium and parallel-line figures	A trapezium with diagonals meeting at O, or a line parallel to its parallel sides; find a segment or prove a ratio.	Use alternate angles and vertically opposite angles to get $△ A O B \sim △ C O D$ , or join a diagonal and apply BPT in two triangles, then combine the ratios.
Assertion-reason on criteria and definitions	An assertion about two figures being similar or a line being parallel, with a reason quoting a criterion or theorem.	Check the assertion with numbers first. Then check if the reason is true in general and whether it is exactly the rule that proves the assertion. Watch for SSA disguised as SAS, and for figures with equal sides but unequal angles.
Real-life case studies	Shadows, lamp posts, mirrors, ramps, ladders or scale maps, usually as a Section E case with 1 + 1 + 2 mark parts.	Draw and label a rough figure, identify the right angles and the shared angle, state AA, then set up the ratio with whole lengths (distance plus shadow for lamp posts).
Prove the Basic Proportionality Theorem	A long answer: state and prove BPT, often followed by a numerical part using it.	Write the statement, given, to prove, figure and construction (join BE, CD; draw perpendiculars). Show the two area ratios, state why $a r (B D E) = a r (D E C)$ , conclude, then solve the application.

Tricks to remember Triangles

Short memory aids. Say each one aloud once, then use it on a question the same day.

Part with part, whole with whole
In BPT and its converse, compare $\frac{top part}{bottom part}$ on both sides, or $\frac{top part}{whole side}$ on both sides. Never mix a part on one side with a whole on the other.
Use it when: Every BPT or converse question, and checking a ratio before you cross-multiply.
Letters sit in matching seats
In $△ A B C \sim △ P QR$ , seat 1 is A and P, seat 2 is B and Q, seat 3 is C and R. Side AB uses seats 1 and 2, so it matches PQ.
Use it when: Writing ratios correctly after proving similarity.
Sort before SSS
Line up sides smallest to largest in both triangles: small with small, middle with middle, large with large. The vertex between two sides matches the vertex between their partners.
Use it when: SSS questions and MCQs where sides are given in a jumbled order.
The A sits between the S's
In SAS, picture the letters: Side, Angle, Side. The angle must be squeezed between the two sides you compared.
Use it when: Avoiding the SSA trap in proofs and assertion-reason items.
The third side needs the whole side
For DE, think of the small triangle ADE against the big triangle ABC: $\frac{D E}{B C} = \frac{A D}{A B}$ , small triangle over big triangle.
Use it when: Finding the length of the parallel line DE.
Quick sanity check
After solving, put your values back: both ratios must be equal, every length positive, and a part shorter than the whole side.
Use it when: Catching wrong roots and arithmetic slips in BPT equations.

Triangles: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: SAS correspondence from an included angle · 1 mark · MCQBoard level

In $△ A B C$ and $△ P QR$ , $\frac{A B}{QR} = \frac{B C}{P R}$ and $∠ B = ∠ R$ . Which of the following is true?
(a) $△ A B C \sim △ P QR$
(b) $△ A B C \sim △ QR P$
(c) $△ A B C \sim △ R P Q$
(d) $△ A B C \sim △ P R Q$

Try it first, then show the solutionSolution

The sides including $∠ B$ are AB and BC. The sides including $∠ R$ are QR and PR.
$∠ B = ∠ R$ and the including sides are proportional, so the triangles are similar by SAS, with $B \leftrightarrow R$ .
AB matches QR, so A matches the other end of QR, which is Q.
BC matches PR, so C matches P.
Correspondence $A \leftrightarrow Q$ , $B \leftrightarrow R$ , $C \leftrightarrow P$ gives $△ A B C \sim △ QR P$ .

Answer: (b) $△ A B C \sim △ QR P$

Example 2: Is two-sides-proportional enough? · 1 mark · Assertion-reasonBoard level

Assertion (A): Any two isosceles right-angled triangles are similar.
Reason (R): If two sides of one triangle are proportional to two sides of another triangle, the triangles are similar.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

An isosceles right triangle has angles $9 0^{\circ}, 4 5^{\circ}, 4 5^{\circ}$ , so any two have equal corresponding angles. By AA they are similar. A is true.
Two proportional sides are not enough: SAS needs the angle between those sides to be equal too.
Example: sides 3 cm and 4 cm with included angle $9 0^{\circ}$ , and sides 3 cm and 4 cm with included angle $6 0^{\circ}$ . Two sides match in ratio $1 : 1$ but the triangles are not similar. R is false.

Answer: (c) A is true but R is false.

Example 3: Two parallel lines inside one triangle · 3 marks · Short answerBoard level

In $△ A B C$ , D is a point on AB and E is a point on AC such that $D E ∥ B C$ . F is a point on BC such that $E F ∥ A B$ . $A D = 4$ cm, $D B = 6$ cm and $B C = 15$ cm. Find DE, BF, FC and EF.

Try it first, then show the solutionSolution

$A B = A D + D B = 10$ cm.
$D E ∥ B C$ , so $∠ A D E = ∠ A B C$ and $∠ A E D = ∠ A C B$ (corresponding angles); hence $△ A D E \sim △ A B C$ (AA).
$\frac{D E}{B C} = \frac{A D}{A B} = \frac{4}{10}$ , so $D E = \frac{4}{10} \times 15 = 6$ cm. [1 mark]
In DBFE, $D E ∥ B F$ and $E F ∥ D B$ , so DBFE is a parallelogram.
Opposite sides are equal: $B F = D E = 6$ cm, so $F C = 15 - 6 = 9$ cm. [1 mark]
Also $E F = D B = 6$ cm.
Check with BPT: in $△ A B C$ , $\frac{C E}{E A} = \frac{D B}{A D} = \frac{6}{4}$ ; in $△ C A B$ with $E F ∥ A B$ , $\frac{C F}{F B} = \frac{9}{6} = \frac{3}{2}$ . Equal, so the answers agree. [1 mark]

Answer: $D E = 6$ cm, $B F = 6$ cm, $F C = 9$ cm, $E F = 6$ cm.

Example 4: Trapezium diagonals with an unknown · 3 marks · Short answerHarder board level

ABCD is a trapezium with $A B ∥ D C$ . Its diagonals AC and BD meet at O. $A O = (3 x - 1)$ cm, $O C = (5 x - 3)$ cm, $B O = (2 x + 1)$ cm and $O D = (6 x - 5)$ cm. Find $x$ , the four lengths, and the ratio $A B : D C$ .

Try it first, then show the solutionSolution

In $△ A O B$ and $△ C O D$ : $∠ O A B = ∠ O C D$ (alternate angles, $A B ∥ D C$ ) and $∠ A O B = ∠ C O D$ (vertically opposite).
So $△ A O B \sim △ C O D$ (AA), giving $\frac{A O}{O C} = \frac{B O}{O D} = \frac{A B}{C D}$ . [1 mark]
$\frac{3 x - 1}{5 x - 3} = \frac{2 x + 1}{6 x - 5}$ , so $(3 x - 1) (6 x - 5) = (5 x - 3) (2 x + 1)$ .
$18 x^{2} - 21 x + 5 = 10 x^{2} - x - 3$ , so $8 x^{2} - 20 x + 8 = 0$ , i.e. $2 x^{2} - 5 x + 2 = 0$ . [1 mark]
$(2 x - 1) (x - 2) = 0$ , so $x = \frac{1}{2}$ or $x = 2$ .
$x = \frac{1}{2}$ gives $O C = - \frac{1}{2}$ , impossible, so $x = 2$ . [½ mark]
$A O = 5$ cm, $O C = 7$ cm, $B O = 5$ cm, $O D = 7$ cm, and $A B : D C = 5 : 7$ . Check: $\frac{5}{7} = \frac{5}{7}$ . [½ mark]

Answer: $x = 2$ ; $A O = 5$ cm, $O C = 7$ cm, $B O = 5$ cm, $O D = 7$ cm; $A B : D C = 5 : 7$ .

Example 5: Prove BPT, then use it in a trapezium · 5 marks · Long answerHarder board level

(i) Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
(ii) ABCD is a trapezium with $A B ∥ D C$ . A line parallel to AB meets AD at E and BC at F. Using (i), show that $\frac{A E}{E D} = \frac{B F}{F C}$ . If $A E = 3$ cm, $E D = 4$ cm and $B C = 14$ cm, find BF and FC.

Try it first, then show the solutionSolution

Given: In $△ A B C$ , $D E ∥ B C$ , with D on AB and E on AC. To prove: $\frac{A D}{D B} = \frac{A E}{E C}$ . Construction: join BE and CD; draw $E N ⊥ A B$ and $D M ⊥ A C$ . [½ mark]
$ar (A D E) = \frac{1}{2} \cdot A D \cdot E N$ and $ar (B D E) = \frac{1}{2} \cdot D B \cdot E N$ , so $\frac{ar ( A D E )}{ar ( B D E )} = \frac{A D}{D B}$ ... (1) [½ mark]
$ar (A D E) = \frac{1}{2} \cdot A E \cdot D M$ and $ar (D E C) = \frac{1}{2} \cdot E C \cdot D M$ , so $\frac{ar ( A D E )}{ar ( D E C )} = \frac{A E}{E C}$ ... (2) [½ mark]
$△ B D E$ and $△ D E C$ are on the same base DE and between the same parallels DE and BC, so $ar (B D E) = ar (D E C)$ ... (3) [1 mark]
From (1), (2) and (3): $\frac{A D}{D B} = \frac{A E}{E C}$ . Proved. [½ mark]
(ii) Join AC; let it meet EF at G. In $△ A D C$ , $E G ∥ D C$ , so $\frac{A E}{E D} = \frac{A G}{GC}$ (BPT). [½ mark]
In $△ C A B$ , $GF ∥ A B$ , so $\frac{C G}{G A} = \frac{C F}{F B}$ , i.e. $\frac{A G}{GC} = \frac{B F}{F C}$ (BPT). Hence $\frac{A E}{E D} = \frac{B F}{F C}$ . [½ mark]
$\frac{B F}{F C} = \frac{3}{4}$ and $B F + F C = 14$ , so $B F = \frac{3}{7} \times 14 = 6$ cm and $F C = 8$ cm. [1 mark]

Answer: BPT proved by comparing areas; $\frac{A E}{E D} = \frac{B F}{F C}$ ; $B F = 6$ cm and $F C = 8$ cm.

Example 6: Crossing wires between two poles · 5 marks · Long answerHarder board level

Two vertical poles AB and CD stand on level ground with feet B and D. $A B = 6$ m, $C D = 9$ m and $B D = 15$ m. Wires AD and CB cross at O. L is the foot of the perpendicular from O to BD.
(i) Prove that $\frac{1}{O L} = \frac{1}{A B} + \frac{1}{C D}$ .
(ii) Find OL and BL.

Try it first, then show the solutionSolution

In $△ B L O$ and $△ B D C$ : $∠ B$ is common and $∠ B L O = ∠ B D C = 9 0^{\circ}$ , so $△ B L O \sim △ B D C$ (AA).
Hence $\frac{O L}{C D} = \frac{B L}{B D}$ ... (1) [1 mark]
In $△ D L O$ and $△ D B A$ : $∠ D$ is common and $∠ D L O = ∠ D B A = 9 0^{\circ}$ , so $△ D L O \sim △ D B A$ (AA).
Hence $\frac{O L}{A B} = \frac{D L}{B D}$ ... (2) [1 mark]
Adding (1) and (2): $\frac{O L}{C D} + \frac{O L}{A B} = \frac{B L + D L}{B D} = \frac{B D}{B D} = 1$ .
Dividing by OL: $\frac{1}{C D} + \frac{1}{A B} = \frac{1}{O L}$ . Proved. [1 mark]
$\frac{1}{O L} = \frac{1}{6} + \frac{1}{9} = \frac{3 + 2}{18} = \frac{5}{18}$ , so $O L = 3.6$ m. [1 mark]
From (1): $B L = \frac{O L \times B D}{C D} = \frac{3.6 \times 15}{9} = 6$ m. Check with (2): $D L = 9$ m and $\frac{3.6}{6} = 0.6 = \frac{9}{15}$ . [1 mark]

Answer: $\frac{1}{O L} = \frac{1}{A B} + \frac{1}{C D}$ ; $O L = 3.6$ m and $B L = 6$ m. (OL does not depend on BD.)

Example 7: Find the error: a wrong use of SAS · 3 marks · Short answerHOTS / competency

Riya writes: "In $△ A B C$ , $A B = 8$ cm, $A C = 5$ cm and $∠ B = 3 0^{\circ}$ . In $△ P QR$ , $P Q = 4$ cm, $P R = 2.5$ cm and $∠ Q = 3 0^{\circ}$ . Since $\frac{A B}{P Q} = \frac{A C}{P R} = 2$ and $∠ B = ∠ Q$ , the triangles are similar by SAS."
(i) Find the error in her reasoning.
(ii) Using the given data, show that the condition she used cannot guarantee similarity. (In a right triangle with a $3 0^{\circ}$ angle, the side opposite $3 0^{\circ}$ is half the hypotenuse. Use Pythagoras as a known result.)
(iii) State one extra piece of information, of each of two kinds, that would make the triangles similar.

Try it first, then show the solutionSolution

(i) In SAS the equal angle must be included between the two proportional sides. The angle between AB and AC is $∠ A$ , not $∠ B$ ; the angle between PQ and PR is $∠ P$ , not $∠ Q$ . So SAS does not apply. [1 mark]
(ii) In $△ A B C$ , draw $A N ⊥ B C$ . In right $△ A N B$ with $∠ B = 3 0^{\circ}$ : $A N = \frac{1}{2} \times 8 = 4$ cm and $B N = 64 - 16 = 43$ cm. C must satisfy $A C = 5$ , so $N C = 25 - 16 = 3$ cm, and C can be 3 cm on either side of N. Since $43 > 3$ , both points lie on the ray from B, so $B C = 43 + 3 \approx 9.93$ cm or $B C = 43 - 3 \approx 3.93$ cm. [½ mark]
In the same way, in $△ P QR$ : $P M = 2$ cm, $QM = 23$ cm, $M R = 6.25 - 4 = 1.5$ cm, so $QR \approx 4.96$ cm or $QR \approx 1.96$ cm. If $B C \approx 9.93$ cm and $QR \approx 1.96$ cm, then $\frac{B C}{QR} \neq = 2$ although $\frac{A B}{P Q} = 2$ , so the triangles are not similar. Two sides and a non-included angle (SSA) can allow two different triangles, so similarity is not guaranteed. [½ mark]
(iii) Kind 1: $∠ A = ∠ P$ . Then SAS applies with $\frac{A B}{P Q} = \frac{A C}{P R}$ , giving $△ A B C \sim △ P QR$ . [½ mark]
Kind 2: $B C = 2 QR$ . Then all three sides are proportional and SSS gives $△ A B C \sim △ P QR$ . [½ mark]

Answer: The angle used is not the included angle, so SAS fails. The data fits two different shapes of each triangle ( $B C \approx 9.93$ cm or $3.93$ cm; $QR \approx 4.96$ cm or $1.96$ cm), so the triangles need not be similar. They become similar if $∠ A = ∠ P$ (SAS) or if $B C = 2 QR$ (SSS).

Example 8: Measuring a tree with a mirror · 3 marks · Short answerBoard level

Meera wants the height of a neem tree in her school. She places a small mirror flat on level ground 12 m from the foot of the tree. She stands 1.5 m from the mirror, on the line joining the mirror and the tree, and just sees the top of the tree in the mirror. Her eyes are 1.2 m above the ground. The angle of incidence equals the angle of reflection, so the lines from her eyes and from the treetop make equal angles with the ground at the mirror.
(i) Find the height of the tree.
(ii) She now stands 2 m from the mirror. How far from the tree must the mirror be for her to see the top again?

Try it first, then show the solutionSolution

Let E be her eyes, G the point below E, M the mirror, T the foot and P the top of the tree. $∠ E GM = ∠ P T M = 9 0^{\circ}$ and $∠ E M G = ∠ P M T$ (equal angles at the mirror).
So $△ E GM \sim △ P T M$ (AA), giving $\frac{E G}{P T} = \frac{GM}{T M}$ . [1 mark]
$\frac{1.2}{P T} = \frac{1.5}{12}$ , so $P T = \frac{1.2 \times 12}{1.5} = 9.6$ m. [1 mark]
(ii) With $GM = 2$ m: $\frac{1.2}{9.6} = \frac{2}{T M}$ , so $T M = \frac{2 \times 9.6}{1.2} = 16$ m. The mirror moves 4 m further from the tree. [1 mark]

Answer: (i) The tree is 9.6 m high. (ii) The mirror must be 16 m from the tree.

Example 9: Prove and deduce: a line through the meeting point of a trapezium's diagonals · 5 marks · Long answerHOTS / competency

ABCD is a trapezium with $A B ∥ D C$ . Its diagonals AC and BD meet at O. A line through O, parallel to AB (and so also parallel to DC), meets AD at P and BC at Q.
(i) Prove that $\frac{A O}{A C} = \frac{B O}{B D}$ .
(ii) Hence prove that O is the midpoint of PQ.
(iii) If $A B = 12$ cm and $D C = 6$ cm, find PQ.

Try it first, then show the solutionSolution

(i) In $△ A O B$ and $△ C O D$ : $∠ O A B = ∠ O C D$ (alternate angles, $A B ∥ D C$ ) and $∠ A O B = ∠ C O D$ (vertically opposite). So $△ A O B \sim △ C O D$ (AA) and $\frac{A O}{O C} = \frac{B O}{O D}$ . [1 mark]
Taking reciprocals and adding 1: $\frac{O C + A O}{A O} = \frac{O D + B O}{B O}$ , i.e. $\frac{A C}{A O} = \frac{B D}{B O}$ , so $\frac{A O}{A C} = \frac{B O}{B D}$ ... (1) [½ mark]
(ii) In $△ A D C$ , $P O ∥ D C$ : $∠ A$ is common and $∠ A P O = ∠ A D C$ (corresponding angles), so $△ A P O \sim △ A D C$ (AA) and $\frac{P O}{D C} = \frac{A O}{A C}$ ... (2) [1 mark]
In $△ B D C$ , $O Q ∥ D C$ : $∠ B$ is common and $∠ B O Q = ∠ B D C$ (corresponding angles), so $△ B O Q \sim △ B D C$ (AA) and $\frac{O Q}{D C} = \frac{B O}{B D}$ ... (3) [1 mark]
From (1), (2) and (3): $\frac{P O}{D C} = \frac{O Q}{D C}$ , so $P O = O Q$ and O is the midpoint of PQ. Proved. [½ mark]
(iii) From (i), $\frac{A O}{O C} = \frac{A B}{C D} = \frac{12}{6} = 2$ , so $\frac{A O}{A C} = \frac{2}{3}$ . [½ mark]
From (2), $P O = \frac{2}{3} \times 6 = 4$ cm, so $P Q = 2 \times 4 = 8$ cm. Check with (3): $\frac{B O}{B D} = \frac{2}{3}$ too, so $O Q = 4$ cm. [½ mark]

Answer: $\frac{A O}{A C} = \frac{B O}{B D}$ ; O is the midpoint of PQ; $P Q = 8$ cm.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Irrigation channels in a triangular field

Gurpreet farms a triangular field ABC near Ludhiana. Side AB is 120 m, side AC is 150 m and side BC, along the road, is 100 m. To water the crop, he digs a straight channel DE parallel to the road BC, starting at D on AB with AD = 48 m and ending at E on AC. Next season he plans a second straight channel FG, also parallel to BC, with F on AB between D and B, and G on AC between E and C. He wants the two boundary stretches between the channels, DF along AB and EG along AC, to total 81 m. Treat the channels as straight line segments.

(i) · 1 mark
Find the length AE.
Answer
Answer: $A E = 60$ m
$D E ∥ B C$ , so by BPT $\frac{A D}{A B} = \frac{A E}{A C}$ , giving $\frac{48}{120} = \frac{A E}{150}$ and $A E = 60$ m.
(ii) · 1 mark
Find the length of channel DE.
Answer
Answer: $D E = 40$ m
$△ A D E \sim △ A B C$ (AA, corresponding angles), so $\frac{D E}{B C} = \frac{A D}{A B} = \frac{2}{5}$ and $D E = \frac{2}{5} \times 100 = 40$ m.
(iii) · 2 marks
Find AF and AG for the second channel, and the length of the second channel FG.
Answer
Answer: $A F = 84$ m, $A G = 105$ m, $F G = 70$ m
Let $A F = x$ m. $F G ∥ B C$ , so by BPT $\frac{A G}{A C} = \frac{A F}{A B}$ , giving $A G = \frac{150}{120} x = \frac{5 x}{4}$ . Then $D F + E G = (x - 48) + (\frac{5 x}{4} - 60) = 81$ , so $\frac{9 x}{4} = 189$ and $x = 84$ [1 mark]. So $A F = 84$ m and $A G = 105$ m (check: $D F = 36$ m, $E G = 45$ m, total 81 m). From $△ A F G \sim △ A B C$ (AA), $F G = \frac{84}{120} \times 100 = 70$ m [1 mark].

Shadow of a mobile tower

A mobile tower AB stands vertically on level ground, with foot B. One afternoon Neha, who is 1.5 m tall, stands upright at point C on the ground, 30 m from B. Let N be the top of her head. She notices that the tip of her shadow falls at exactly the same point E as the tip of the tower's shadow, with B, C and E in a straight line and CE = 2 m. Take the sun's rays as parallel and the ground as horizontal.

(i) · 1 mark
Name the two similar triangles in this situation and the criterion that makes them similar.
Answer
Answer: $△ E C N \sim △ E B A$ (AA)
$∠ E C N = ∠ E B A = 9 0^{\circ}$ (Neha and the tower are both vertical) and $∠ E$ is common, so $△ E C N \sim △ E B A$ by AA similarity.
(ii) · 1 mark
Find the height of the tower.
Answer
Answer: $A B = 24$ m
$\frac{A B}{N C} = \frac{E B}{E C}$ , so $\frac{A B}{1.5} = \frac{30 + 2}{2} = 16$ and $A B = 24$ m.
(iii) · 2 marks
An hour later the sun is lower and Neha's shadow is 2.5 m long. Prove that the triangle formed by Neha and her new shadow is similar to the triangle formed by the tower and its new shadow, and find how much longer the tower's shadow has become.
Answer
Answer: The tower's shadow becomes 40 m, which is 8 m longer.
Both stand vertically, so each triangle has a $9 0^{\circ}$ angle at the foot. The sun's rays are parallel, so they make equal angles with the ground at the tip of each shadow. Hence the two triangles are similar (AA) [1 mark]. If the tower's new shadow is $s$ m, then $\frac{24}{1.5} = \frac{s}{2.5}$ , so $s = 16 \times 2.5 = 40$ m. Its earlier shadow was $B E = 32$ m, so it is $40 - 32 = 8$ m longer [1 mark].

Scale drawing of a school garden

A triangular garden ABC is to be made in a school. Its sides are AB = 45 m, BC = 60 m and CA = 75 m. The art teacher makes a scale drawing PQR of the garden on chart paper, with PQ = 6 cm, QR = 8 cm and RP = 10 cm, where P, Q and R stand for A, B and C. On the drawing she marks a straight path ST, with S on PQ and T on PR, PS = 2.4 cm and PT = 4 cm.

(i) · 1 mark
Show that the drawing is similar to the garden and find the scale of the drawing.
Answer
Answer: $△ P QR \sim △ A B C$ (SSS); 1 cm represents 7.5 m
$\frac{A B}{P Q} = \frac{45}{6} = 7.5$ , $\frac{B C}{QR} = \frac{60}{8} = 7.5$ , $\frac{C A}{R P} = \frac{75}{10} = 7.5$ . All three ratios are equal, so the triangles are similar by SSS, and 1 cm on the drawing represents 7.5 m.
(ii) · 1 mark
Find the actual perimeter of the garden using the drawing.
Answer
Answer: 180 m
Perimeter of the drawing $= 6 + 8 + 10 = 24$ cm. Perimeters are in the same ratio as sides, so the actual perimeter $= 24 \times 7.5 = 180$ m. Check: $45 + 60 + 75 = 180$ m.
(iii) · 2 marks
Prove that the path ST is parallel to QR and that $△ P S T \sim △ P QR$ . Hence find the actual length of the path in the garden.
Answer
Answer: $S T ∥ QR$ , $△ P S T \sim △ P QR$ ; the path is 24 m long
$S Q = 6 - 2.4 = 3.6$ cm and $T R = 10 - 4 = 6$ cm. $\frac{P S}{S Q} = \frac{2.4}{3.6} = \frac{2}{3}$ and $\frac{P T}{T R} = \frac{4}{6} = \frac{2}{3}$ , so $S T ∥ QR$ (converse of BPT). Then $∠ P$ is common and $∠ P S T = ∠ P QR$ (corresponding angles), so $△ P S T \sim △ P QR$ (AA) [1 mark]. $\frac{S T}{QR} = \frac{P S}{P Q} = \frac{2.4}{6} = \frac{2}{5}$ , so $S T = \frac{2}{5} \times 8 = 3.2$ cm on the drawing, and the actual path is $3.2 \times 7.5 = 24$ m [1 mark].

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

In $△ A B C$ , D and E are points on AB and AC such that $D E ∥ B C$ . If $A D = 2.4$ cm, $A E = 3.2$ cm and $E C = 4.8$ cm, find AB. (2 marks)

Step	Marks
Since $D E ∥ B C$ , by BPT $\frac{A D}{D B} = \frac{A E}{E C}$ , so $\frac{2.4}{D B} = \frac{3.2}{4.8}$	1
$D B = \frac{2.4 \times 4.8}{3.2} = 3.6$ cm	0.5
$A B = A D + D B = 2.4 + 3.6 = 6$ cm	0.5

In $△ A B C$ , P and Q are points on AB and AC such that $A P = 3$ cm, $P B = 6$ cm, $A Q = 5$ cm and $QC = 10$ cm. Show that $P Q ∥ B C$ . If $B C = 12$ cm, find PQ. (3 marks)

Step	Marks
$\frac{A P}{P B} = \frac{3}{6} = \frac{1}{2}$	0.5
$\frac{A Q}{QC} = \frac{5}{10} = \frac{1}{2}$	0.5
Since $\frac{A P}{P B} = \frac{A Q}{QC}$ , $P Q ∥ B C$ (converse of BPT)	0.5
$∠ A$ common and $∠ A P Q = ∠ A B C$ (corresponding angles), so $△ A P Q \sim △ A B C$ (AA)	1
$\frac{P Q}{B C} = \frac{A P}{A B} = \frac{3}{9}$ , so $P Q = \frac{1}{3} \times 12 = 4$ cm	0.5

State and prove the Basic Proportionality Theorem. Using it, find $x$ if in $△ A B C$ , $D E ∥ B C$ with D on AB and E on AC, $A D = (x + 1)$ cm, $D B = (3 x - 1)$ cm, $A E = (x + 3)$ cm and $E C = (3 x + 4)$ cm. (5 marks)

Step	Marks
Statement; given $△ A B C$ with $D E ∥ B C$ ; to prove $\frac{A D}{D B} = \frac{A E}{E C}$ ; correct figure	1
Construction: join BE and CD; draw $E N ⊥ A B$ and $D M ⊥ A C$	0.5
$\frac{a r ( A D E )}{a r ( B D E )} = \frac{\frac{1}{2} A D \cdot E N}{\frac{1}{2} D B \cdot E N} = \frac{A D}{D B}$ and $\frac{a r ( A D E )}{a r ( D E C )} = \frac{A E}{E C}$	1
$a r (B D E) = a r (D E C)$ (same base DE, between the same parallels DE and BC)	1
Hence $\frac{A D}{D B} = \frac{A E}{E C}$	0.5
$\frac{x + 1}{3 x - 1} = \frac{x + 3}{3 x + 4}$ gives $3 x^{2} + 7 x + 4 = 3 x^{2} + 8 x - 3$ , so $x = 7$ (check: $\frac{8}{20} = \frac{10}{25}$ )	1

Where students lose marks in Triangles

The slips examiners see most often, and what to write instead.

Mistake: Using $\frac{A D}{D B}$ to find the parallel side DE
Students remember the BPT ratio and apply it to the third side as well, but BPT says nothing about DE.
Wrong working
$A D = 2$ cm, $D B = 3$ cm, $B C = 10$ cm. $\frac{D E}{B C} = \frac{A D}{D B} = \frac{2}{3}$ , so $D E = \frac{20}{3}$ cm.
Right working
$△ A D E \sim △ A B C$ (AA, corresponding angles as $D E ∥ B C$ ). $\frac{D E}{B C} = \frac{A D}{A B} = \frac{2}{5}$ , so $D E = 4$ cm.
Do this: For DE, always prove the similarity and use the whole side AB in the denominator.
Usually costs: Usually 1 to 2 marks, as the final answer and the method step are both wrong.
Mistake: Writing the similarity statement in the wrong order
Students write vertices alphabetically instead of by correspondence, especially when sides are jumbled.
Wrong working
$A B = 6, B C = 9, C A = 12$ ; $D E = 16, E F = 8, F D = 12$ . So $△ A B C \sim △ D E F$ and $∠ C = ∠ F$ .
Right working
Matching sides: $A B \leftrightarrow E F$ , $B C \leftrightarrow F D$ , $C A \leftrightarrow D E$ . So $△ A B C \sim △ E F D$ and $∠ C = ∠ D$ .
Do this: Match sides by size first, then match vertices by the pair of sides that meet there.
Usually costs: The whole 1 mark in an MCQ, or ½ to 1 mark in a longer answer.
Mistake: Using SSA as if it were SAS
The angle is equal and two sides are proportional, so it looks like SAS, but the angle is not between those sides.
Wrong working
$\frac{A B}{P Q} = \frac{A C}{P R}$ and $∠ B = ∠ Q$ , so $△ A B C \sim △ P QR$ (SAS).
Right working
For SAS with sides AB, AC and PQ, PR, the included angles are $∠ A$ and $∠ P$ . You need $∠ A = ∠ P$ ; $∠ B = ∠ Q$ does not give similarity here.
Do this: Check that the equal angle is formed by the two sides you compared.
Usually costs: The similarity step and everything after it, often 2 or more marks.
Mistake: Giving no reasons in a similarity proof
Students see the equal angles in the figure and skip writing why they are equal.
Wrong working
$∠ A = ∠ A$ , $∠ A D E = ∠ A B C$ . So the triangles are similar.
Right working
$∠ D A E = ∠ B A C$ (common). $∠ A D E = ∠ A B C$ (corresponding angles, $D E ∥ B C$ ). So $△ A D E \sim △ A B C$ (AA similarity).
Do this: Give every equality a reason in brackets and name the criterion at the end.
Usually costs: ½ to 1 mark in a 3-mark or 5-mark proof.
Mistake: Comparing a part with a whole in the converse of BPT
Given lengths include a whole side, and students divide by it on one side only.
Wrong working
$A D = 3, D B = 6, A E = 4, A C = 12$ . $\frac{A D}{D B} = \frac{1}{2}$ , $\frac{A E}{A C} = \frac{1}{3}$ , not equal, so DE is not parallel to BC.
Right working
$E C = 12 - 4 = 8$ . $\frac{A D}{D B} = \frac{3}{6} = \frac{1}{2}$ , $\frac{A E}{E C} = \frac{4}{8} = \frac{1}{2}$ . Equal, so $D E ∥ B C$ (converse of BPT).
Do this: Find the missing part first and compare part with part on both sides.
Usually costs: Usually the full question, since the conclusion flips.
Mistake: Skipping the reason for equal areas in the BPT proof
Students memorise the lines of the proof but not why the key step is true.
Wrong working
$a r (B D E) = a r (D E C)$ , so $\frac{A D}{D B} = \frac{A E}{E C}$ .
Right working
$a r (B D E) = a r (D E C)$ (triangles on the same base DE and between the same parallels DE and BC). Hence $\frac{A D}{D B} = \frac{A E}{E C}$ .
Do this: Learn the proof as reasons, not just lines. Also include the construction: join BE and CD, draw the perpendiculars.
Usually costs: About 1 mark of the 5-mark proof; missing construction costs another ½.
Mistake: Using only the distance instead of distance plus shadow in lamp-post problems
Students pair the lamp post with the distance to the person rather than the full base of the big triangle.
Wrong working
$\frac{1.6}{4.8} = \frac{s}{6}$ , so $s = 2$ m.
Right working
$\frac{1.6}{4.8} = \frac{s}{s + 6}$ , so $s + 6 = 3 s$ and $s = 3$ m.
Do this: Draw the figure and mark the big triangle from the top of the post to the tip of the shadow.
Usually costs: 1 to 2 marks in a case-study part.

NCERT exercise guide: Triangles

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise 6.1

What it tests: The meaning of similar figures. You fill in blanks about when figures are similar, give examples of pairs that are always similar, and give counter examples of pairs that need not be similar.

How to approach it: Recall that two polygons with the same number of sides are similar only when both conditions hold: corresponding angles are equal and corresponding sides are in the same ratio. One condition alone is not enough. To show a pair need not be similar, describe one pair that meets one condition but fails the other.

Like the always-similar and counter example questions in Exercise 6.1

State whether each pair is always similar. Give a reason or a counter example.
(i) Two regular hexagons
(ii) A square and a rhombus
(iii) Two right-angled triangles

Show the solutionSolution

(i) Each interior angle of every regular hexagon is $12 0^{\circ}$ , so corresponding angles are equal.
All six sides of a regular hexagon are equal, so for hexagons of sides $a$ and $b$ every pair of corresponding sides is in the ratio $a : b$ .
Both conditions hold, so two regular hexagons are always similar.
(ii) A square of side 3 cm and a rhombus of side 3 cm with angles $6 0^{\circ}$ and $12 0^{\circ}$ have sides in the ratio $1 : 1$ .
But the square has angles of $9 0^{\circ}$ and the rhombus does not, so the angles are not equal. They are not similar.
(iii) A triangle with angles $9 0^{\circ}, 4 5^{\circ}, 4 5^{\circ}$ and one with angles $9 0^{\circ}, 3 0^{\circ}, 6 0^{\circ}$ are both right-angled.
Their other angles are not equal, so they are not similar.

Answer: (i) Always similar. (ii) Not always similar: equal side ratios but the angles can differ. (iii) Not always similar: the acute angles can differ.

Exercise 6.2

What it tests: The Basic Proportionality Theorem and its converse. You find unknown lengths when a line is parallel to one side, decide whether a line is parallel from given lengths, and prove results by applying the theorem in two triangles.

How to approach it: Mark the parallel line and name the triangle it cuts. Write the BPT ratio with the vertex first, such as $\frac{A D}{D B} = \frac{A E}{E C}$ . For a proof, apply BPT in one triangle, then in a second triangle that shares a ratio, and equate. To prove a line parallel, show the two ratios are equal and quote the converse of BPT.

Like the proofs that use BPT in two triangles in Exercise 6.2

O is a point inside $△ P QR$ . A, B and C are points on OP, OQ and OR respectively such that $A B ∥ P Q$ and $A C ∥ P R$ . $O A = 2$ cm, $A P = 3$ cm, $O C = 4$ cm and $O B = 3$ cm.
(i) Find CR and BQ.
(ii) Prove that $B C ∥ QR$ .

Show the solutionSolution

In $△ O P Q$ , $A B ∥ P Q$ , so by BPT $\frac{O A}{A P} = \frac{O B}{B Q}$ ... (1)
In $△ O P R$ , $A C ∥ P R$ , so by BPT $\frac{O A}{A P} = \frac{O C}{C R}$ ... (2)
From (2): $\frac{2}{3} = \frac{4}{C R}$ , so $C R = 6$ cm.
From (1): $\frac{2}{3} = \frac{3}{B Q}$ , so $B Q = 4.5$ cm.
From (1) and (2): $\frac{O B}{B Q} = \frac{O C}{C R}$ .
In $△ O QR$ , B and C divide OQ and OR in the same ratio, so by the converse of BPT, $B C ∥ QR$ .
Check: $\frac{3}{4.5} = \frac{2}{3}$ and $\frac{4}{6} = \frac{2}{3}$ , equal.

Answer: $C R = 6$ cm, $B Q = 4.5$ cm, and $B C ∥ QR$ by the converse of BPT.

Exercise 6.3

What it tests: The AA, SSS and SAS similarity criteria. You decide which pairs of triangles are similar, write the correct correspondence, find unknown sides or angles, and prove product relations such as $P A \cdot P B = P C \cdot P D$ .

How to approach it: Look for a common angle, vertically opposite angles, right angles, or angles made by parallel lines. Two equal angles give AA. For a product relation, rewrite it as a ratio, find the two triangles containing those four segments, prove them similar, then cross-multiply. Always match vertices in the order of the equal angles.

Like the product-relation proofs with perpendiculars in Exercise 6.3

In $△ P QR$ , $∠ Q$ is acute. M is the foot of the perpendicular from P to QR and N is the foot of the perpendicular from R to PQ. Prove that $QM \cdot QR = QN \cdot QP$ . If $QR = 10$ cm, $QM = 4$ cm and $QP = 8$ cm, find QN.

Show the solutionSolution

Consider $△ QM P$ and $△ QN R$ .
$∠ QM P = ∠ QN R = 9 0^{\circ}$ (given perpendiculars).
$∠ M QP = ∠ N QR$ (common angle Q).
So $△ QM P \sim △ QN R$ (AA criterion).
Corresponding sides are proportional: $\frac{QM}{QN} = \frac{QP}{QR}$ .
Cross-multiplying: $QM \cdot QR = QN \cdot QP$ . Proved.
Substitute: $4 \times 10 = QN \times 8$ , so $QN = 5$ cm.

Answer: $QM \cdot QR = QN \cdot QP$ (from $△ QM P \sim △ QN R$ ), and $QN = 5$ cm.

Triangles: previous year questions

Standard: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 8.2 marks per paper; 93 questions in all, counting OR alternatives. In 4 of 4 sample papers.
Basic: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 8.4 marks per paper; 90 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Triangles

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026 Basic, set 430/1/1, Q8 · 1 mark · MCQ

From the figures given below, which of the following is true about the measure of $∠ P$ ?

(A) $∠ P = 6 0^{\circ}$
(B) $∠ P = 8 0^{\circ}$
(C) $∠ P = 4 0^{\circ}$
(D)The measure of $∠ P$ cannot be determined

Figure in the paper: Two triangles. Triangle ABC: AB = 3.8 cm, BC = 6 cm, AC = $33$ cm, angle at A = 80°, angle at B = 60°. Triangle PQR: PR = $63$ cm, RQ = 7.6 cm, PQ = 12 cm; the angle at P is marked as the angle to be determined. See it in the official paper.

Try it first, then show our solutionSolution

Compare sides: $\frac{R Q}{A B} = \frac{7.6}{3.8} = 2$ , $\frac{P Q}{B C} = \frac{12}{6} = 2$ , $\frac{P R}{A C} = \frac{6 3}{3 3} = 2$ . So $△ A B C \sim △ R QP$ [SSS similarity], which gives $∠ P = ∠ C$ .
½ mark
$∠ C = 18 0^{\circ} - 8 0^{\circ} - 6 0^{\circ} = 4 0^{\circ}$ , so $∠ P = 4 0^{\circ}$ .
½ mark

Answer: ∠P = 40°, option (C)

Where the marks are

Uses SSS similarity with the correct vertex correspondence (½ mark)
Finds $∠ P = 4 0^{\circ}$ (½ mark)

See it in the full 2026 board paper, set 430/1/1

2026 Basic, set 430/1/1, Q22(a) · 2 marks · Very short answer

In the given figure, if $P Q ∥ R S$ , then prove that $△ P O Q \sim △ S O R$ .

Figure in the paper: Lines PQ and RS are parallel (arrow marks). Segments PS and QR cross at O; P and Q are on one side, R and S on the other, with P-O-S and Q-O-R. See it in the official paper.

Try it first, then show our solutionSolution

In $△ P O Q$ and $△ S O R$ : $P Q ∥ R S$ and $P S$ is a transversal, so $∠ O P Q = ∠ O S R$ [alternate interior angles].
1 mark
$∠ P O Q = ∠ S O R$ [vertically opposite angles]. (Also $∠ O QP = ∠ O R S$ , alternate angles with transversal $QR$ .)
½ mark
Hence $△ P O Q \sim △ S O R$ [AA similarity criterion].
½ mark

Answer: $△ P O Q \sim △ S O R$ by AA similarity

Where the marks are

Uses alternate interior angles $∠ O P Q = ∠ O S R$ (1 mark)
Uses vertically opposite angles $∠ P O Q = ∠ S O R$ (½ mark)
Concludes similarity by AA criterion (½ mark)

See it in the full 2026 board paper, set 430/1/1

2026 Basic, set 430/1/1, Q33 · 5 marks · Long answer

State "Basic Proportionality Theorem" and use it to prove the following : In a quadrilateral ABCD, diagonals AC and BD intersect each other at O such that $\frac{A O}{B O} = \frac{C O}{D O}$ as shown in the given figure. Prove that ABCD is a trapezium.

Figure in the paper: Quadrilateral ABCD with diagonals AC and BD meeting at O. Vertices in order: A bottom left, B bottom right, C top right, D top left. See it in the official paper.

Try it first, then show our solutionSolution

Basic Proportionality Theorem: If a line is drawn parallel to one side of a triangle so that it meets the other two sides at distinct points, then it divides those two sides in the same ratio.
1 mark
Given: ABCD is a quadrilateral whose diagonals AC and BD meet at O, with $\frac{A O}{B O} = \frac{C O}{D O}$ . To prove: ABCD is a trapezium. Construction: Through O, draw $O E ∥ A B$ meeting AD at E. (Figure: quadrilateral ABCD with diagonals meeting at O and the line OE drawn parallel to AB.)
1 mark
In $△ A B D$ , $E O ∥ A B$ , so by BPT: $\frac{D E}{E A} = \frac{D O}{O B}$ …(1)
1 mark
Given $\frac{A O}{B O} = \frac{C O}{D O} \Rightarrow \frac{A O}{C O} = \frac{B O}{D O} \Rightarrow \frac{C O}{A O} = \frac{D O}{B O}$ …(2). From (1) and (2): $\frac{D E}{E A} = \frac{C O}{O A}$ .
1 mark
In $△ A D C$ , E and O divide sides AD and AC in the same ratio, so $E O ∥ D C$ [converse of BPT]. Since $E O ∥ A B$ and $E O ∥ D C$ , we get $A B ∥ D C$ . Hence ABCD has a pair of parallel opposite sides, so ABCD is a trapezium.
1 mark

Answer: BPT stated. $A B ∥ D C$ , so ABCD is a trapezium.

Where the marks are

Correct statement of BPT (1 mark)
Given, to prove, construction $O E ∥ A B$ , with figure (1 mark)
Applies BPT in $△ A B D$ : $\frac{D E}{E A} = \frac{D O}{O B}$ (1 mark)
Uses the given ratio to get $\frac{D E}{E A} = \frac{C O}{O A}$ (1 mark)
Converse of BPT gives $E O ∥ D C$ , hence $A B ∥ D C$ and ABCD is a trapezium (1 mark)

See it in the full 2026 board paper, set 430/1/1

Every Triangles question, by year

Standard (106 questions, sample papers included)

For 2027

Sample paper 2026-27: Q6 (1 mark, solved), Q22(A) (2 marks, solved), Q22(B) (2 marks, solved), Q33 (5 marks, solved)

2026

Set 30/1/1: Q7 (1 mark, solved), Q22(A) (2 marks, solved), Q22(B) (2 marks, solved), Q34(A) (5 marks, solved), Q34(B) (5 marks, solved)

Set 30/2/1: Q8 (1 mark, solved), Q22(a) (2 marks, solved), Q22(b) (2 marks, solved), Q34(a) (5 marks, solved)

Set 30/3/1: Q3 (1 mark, solved), Q13 (1 mark, solved), Q24 (2 marks, solved), Q33 (5 marks, solved)

Set 30/4/1: Q2 (1 mark, solved), Q6 (1 mark, solved), Q22 (2 marks, solved), Q36 (4 marks, solved)

Set 30/5/1: Q5 (1 mark, solved), Q15 (1 mark, solved), Q23 (2 marks, solved), Q33(a) (5 marks, solved), Q33(b) (5 marks, solved)

Sample paper 2025-26: Q14 (1 mark, solved), Q23 (2 marks, solved), Q33 (5 marks, solved)

2025

Set 30/1/1: Q8 (1 mark, solved), Q24(a) (2 marks, solved), Q24(b) (2 marks, solved), Q33(a) (5 marks, solved), Q33(b) (5 marks, solved)

Set 30/2/1: Q8 (1 mark, solved), Q9 (1 mark, solved), Q24(a) (2 marks, solved), Q24(b) (2 marks, solved), Q34 (5 marks, solved)

Set 30/3/1: Q10 (1 mark, solved), Q11 (1 mark, solved), Q18 (1 mark, solved), Q32 (5 marks, solved), Q34(a) (5 marks, solved), Q34(b) (5 marks, solved)

Set 30/4/1: Q7 (1 mark, solved), Q16 (1 mark, solved), Q22(B) (2 marks, solved), Q33(A) (5 marks, solved), Q33(B) (5 marks, solved)

Set 30/5/1: Q8 (1 mark, solved), Q9 (1 mark, solved), Q24 (2 marks, solved), Q33 (5 marks, solved)

Set 30/6/1: Q6 (1 mark, solved), Q7 (1 mark, solved), Q24 (2 marks, solved), Q33(a) (5 marks, solved), Q33(b) (5 marks, solved)

Sample paper 2024-25: Q26(A) (3 marks, solved), Q26(B) (3 marks, solved), Q37 (4 marks, solved)

2024

Set 30/1/1: Q24 (2 marks, solved), Q34(A) (5 marks, solved), Q34(B) (5 marks, solved)

Set 30/2/1: Q6 (1 mark, solved), Q10 (1 mark, solved), Q11 (1 mark, solved), Q33(a) (5 marks, solved), Q33(b) (5 marks, solved)

Set 30/3/1: Q11 (1 mark, solved), Q12 (1 mark, solved), Q24 (2 marks, solved), Q33 (5 marks, solved)

Set 30/4/1: Q13 (1 mark, solved), Q25 (2 marks, solved), Q33(a) (5 marks, solved), Q33(b) (5 marks, solved)

Set 30/5/1: Q17 (1 mark, solved), Q19 (1 mark, solved), Q34(a) (5 marks, solved), Q34(b) (5 marks, solved)

Sample paper 2023-24: Q8 (1 mark, solved), Q22 (2 marks, solved), Q33 (5 marks, solved)

2023

Set 30/1/1: Q22 (2 marks, solved), Q32 (5 marks, solved)

Set 30/2/1: Q3 (1 mark, solved), Q12 (1 mark, solved), Q18 (1 mark, solved), Q35(a) (5 marks, solved), Q35(b) (5 marks, solved)

Set 30/4/1: Q12 (1 mark, solved), Q16 (1 mark, solved), Q33(A) (5 marks, solved), Q33(B) (5 marks, solved)

Set 30/5/1: Q4 (1 mark, solved), Q8 (1 mark, solved), Q13 (1 mark, solved), Q28(a) (3 marks, solved), Q28(b) (3 marks, solved)

Set 30/6/1: Q8 (1 mark, solved), Q10 (1 mark, solved), Q25(A) (2 marks, solved), Q25(B) (2 marks, solved), Q28 (3 marks, solved)

Basic (107 questions, sample papers included)

For 2027

Sample paper 2026-27: Q15 (1 mark, solved), Q16 (1 mark, solved), Q20 (1 mark, solved), Q32 (5 marks, solved)

2026

Set 430/1/1: Q7 (1 mark), Q8 (1 mark, solved), Q22(a) (2 marks, solved), Q22(b) (2 marks, solved), Q33 (5 marks, solved)

Set 430/2/1: Q8 (1 mark, solved), Q20 (1 mark, solved), Q22 (2 marks, solved), Q33 (5 marks, solved)

Set 430/3/1: Q9 (1 mark, solved), Q10 (1 mark, solved), Q22(a) (2 marks, solved), Q22(b) (2 marks, solved), Q33(a) (5 marks, solved), Q33(b) (5 marks, solved)

Set 430/4/1: Q15 (1 mark, solved), Q25 (2 marks, solved), Q33(a) (5 marks, solved), Q33(b) (5 marks, solved)

Set 430/5/1: Q5 (1 mark, solved), Q11 (1 mark, solved), Q25 (2 marks, solved), Q32(a) (5 marks, solved), Q32(b) (5 marks, solved)

Sample paper 2025-26: Q9 (1 mark, solved), Q10 (1 mark, solved), Q19 (1 mark, solved), Q32 (5 marks, solved)

2025

Set 430/1/1: Q7 (1 mark), Q8 (1 mark, solved), Q22(a) (2 marks, solved), Q22(b) (2 marks, solved), Q33 (5 marks, solved)

Set 430/2/1: Q8 (1 mark, solved), Q20 (1 mark, solved), Q22 (2 marks, solved), Q33 (5 marks, solved)

Set 430/3/1: Q9 (1 mark, solved), Q10 (1 mark, solved), Q22(b) (2 marks, solved), Q33(a) (5 marks, solved), Q33(b) (5 marks, solved)

Set 430/4/1: Q18 (1 mark, solved), Q25 (2 marks, solved), Q34(a) (5 marks, solved), Q34(b) (5 marks, solved)

Set 430/5/1: Q31 (3 marks, solved), Q33(a) (5 marks, solved), Q33(b) (5 marks, solved), Q36 (4 marks, solved)

Set 430/6/1: Q11 (1 mark, solved), Q20 (1 mark), Q22 (2 marks, solved), Q36 (4 marks, solved)

Sample paper 2024-25: Q6 (1 mark, solved), Q8 (1 mark, solved), Q11 (1 mark, solved), Q33 (5 marks, solved)

2024

Set 430/1/1: Q21 (2 marks, solved), Q32(A) (5 marks, solved), Q32(B) (5 marks, solved)

Set 430/2/1: Q19 (1 mark, solved), Q24(a) (2 marks, solved), Q24(b) (2 marks, solved), Q34(a) (5 marks, solved), Q34(b) (5 marks, solved)

Set 430/3/1: Q6 (1 mark, solved), Q30 (3 marks, solved), Q38 (4 marks, solved)

Set 430/4/1: Q18 (1 mark, solved), Q23(a) (2 marks, solved), Q23(b) (2 marks, solved), Q34(a) (5 marks, solved), Q34(b) (5 marks, solved)

Set 430/5/1: Q17 (1 mark, solved), Q18 (1 mark, solved), Q34(a) (5 marks, solved), Q34(b) (5 marks, solved)

Sample paper 2023-24: Q7 (1 mark, solved), Q8 (1 mark, solved), Q22(a) (2 marks, solved), Q22(b) (2 marks, solved), Q33 (5 marks, solved)

2023

Set 430/1/1: Q8 (1 mark, solved), Q9 (1 mark, solved), Q15 (1 mark, solved), Q16 (1 mark, solved), Q25 (2 marks, solved), Q31(a) (3 marks, solved), Q31(b) (3 marks, solved)

Set 430/2/1: Q21 (2 marks, solved), Q34 (5 marks, solved)

Set 430/4/1: Q8 (1 mark, solved), Q25 (2 marks, solved), Q38 (4 marks, solved)

Set 430/5/1: Q9 (1 mark, solved), Q13 (1 mark, solved), Q17 (1 mark, solved), Q34 (5 marks, solved)

Set 430/6/1: Q28(a) (3 marks, solved), Q28(b) (3 marks, solved), Q35(a) (5 marks, solved), Q35(b) (5 marks, solved)

Practice questions on Triangles

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
In $△ A B C$ and $△ D E F$ , $∠ A = ∠ F$ and $∠ B = ∠ E$ . Which of the following is correct?
(a) $\frac{A B}{D E} = \frac{B C}{E F}$
(b) $\frac{A B}{F E} = \frac{B C}{E D}$
(c) $\frac{A B}{D F} = \frac{B C}{F E}$
(d) $\frac{A C}{E F} = \frac{B C}{D E}$
Hint
Write the correspondence from the equal angles before you touch the sides.
Answer
(b) $\frac{A B}{F E} = \frac{B C}{E D}$
$A \leftrightarrow F$ , $B \leftrightarrow E$ , so $C \leftrightarrow D$ and $△ A B C \sim △ F E D$ (AA). Hence $\frac{A B}{F E} = \frac{B C}{E D} = \frac{C A}{D F}$ .
Q2 · 1 mark · Assertion-reasonBoard level
Assertion (A): In $△ A B C$ , D is on AB and E is on AC with $D E ∥ B C$ . If $A D = 3$ cm, $D B = 2$ cm and $A E = 4.5$ cm, then $E C = 3$ cm.
Reason (R): A line drawn parallel to one side of a triangle, meeting the other two sides in distinct points, divides those sides in the same ratio.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
R is the Basic Proportionality Theorem; use it to test A.
Answer
(a) Both A and R are true and R is the correct explanation of A.
By BPT, $\frac{3}{2} = \frac{4.5}{E C}$ , so $E C = 3$ cm, which is exactly what A says.
Q3 · 2 marks · Very short answerBoard level
Give a counter example to show that two quadrilaterals with all corresponding angles equal need not be similar. Then state what extra condition would make them similar.
Hint
Think of two quadrilaterals where every angle is $9 0^{\circ}$ .
Answer
A square of side 2 cm and a rectangle 2 cm by 4 cm both have all angles $9 0^{\circ}$ , but their sides are in ratios $\frac{2}{2} = 1$ and $\frac{2}{4} = \frac{1}{2}$ , which differ, so they are not similar.
They would be similar if, in addition, all corresponding sides were in the same ratio.
Q4 · 3 marks · Short answerHarder board level
In $△ A B C$ , D is a point on AB and E is a point on AC such that $A D = 4$ cm, $D B = 6$ cm, $A E = 6$ cm and $E C = 9$ cm. BE and CD meet at O.
(i) Prove that $D E ∥ B C$ .
(ii) Prove that $△ O D E \sim △ O C B$ .
(iii) If $C D = 14$ cm, find OD and OC.
Hint
Use the converse of BPT first; then the parallel lines give alternate angles for the triangles meeting at O. You need $\frac{D E}{B C}$ , which comes from $△ A D E \sim △ A B C$ .
Answer
(i) $\frac{A D}{D B} = \frac{4}{6} = \frac{2}{3}$ and $\frac{A E}{E C} = \frac{6}{9} = \frac{2}{3}$ , so $D E ∥ B C$ (converse of BPT) [1 mark].
(ii) $∠ O D E = ∠ O C B$ (alternate angles, $D E ∥ B C$ , transversal DC) and $∠ D O E = ∠ C O B$ (vertically opposite), so $△ O D E \sim △ O C B$ (AA) [1 mark].
(iii) $△ A D E \sim △ A B C$ (AA, corresponding angles), so $\frac{D E}{B C} = \frac{A D}{A B} = \frac{4}{10} = \frac{2}{5}$ . From (ii), $\frac{O D}{O C} = \frac{D E}{C B} = \frac{2}{5}$ , so $O D = \frac{2}{7} \times 14 = 4$ cm and $O C = 10$ cm [1 mark].
Q5 · 5 marks · Long answerHarder board level
State and prove the Basic Proportionality Theorem. Using it, prove that the line drawn through the midpoint of one side of a triangle, parallel to a second side, bisects the third side.
Hint
In the proof, join BE and CD, drop perpendiculars, and use the fact that triangles on the same base between the same parallels have equal areas.
Answer
Proof [3 marks]: with $D E ∥ B C$ , $\frac{ar ( A D E )}{ar ( B D E )} = \frac{A D}{D B}$ and $\frac{ar ( A D E )}{ar ( D E C )} = \frac{A E}{E C}$ ; $ar (B D E) = ar (D E C)$ (same base DE, same parallels), so $\frac{A D}{D B} = \frac{A E}{E C}$ .
Application [2 marks]: D is the midpoint of AB and $D E ∥ B C$ , so $\frac{A E}{E C} = \frac{A D}{D B} = 1$ . Hence $A E = E C$ and E is the midpoint of AC.
Q6 · 5 marks · Long answerHarder board level
ABCD is a trapezium with $A B ∥ D C$ , $A B = 10$ cm and $D C = 24$ cm. E is a point on AD and F is a point on BC such that $E F ∥ A B$ and $A E : E D = 3 : 4$ .
(i) Join AC, meeting EF at G. Prove that $\frac{A G}{GC} = \frac{3}{4}$ .
(ii) Find EG, GF and hence EF.
(iii) Where must E be placed on AD so that EF is 17 cm long?
Hint
EF is not one side of a single triangle. Split it at G: EG lies in $△ A D C$ and GF lies in $△ C A B$ .
Answer
(i) $E F ∥ A B ∥ D C$ , so in $△ A D C$ , $E G ∥ D C$ and by BPT $\frac{A G}{GC} = \frac{A E}{E D} = \frac{3}{4}$ [1 mark].
(ii) $△ A E G \sim △ A D C$ (AA, corresponding angles), so $\frac{E G}{D C} = \frac{A E}{A D} = \frac{3}{7}$ and $E G = \frac{3}{7} \times 24 = \frac{72}{7}$ cm [1 mark].
From (i), $\frac{C G}{C A} = \frac{4}{7}$ . In $△ C A B$ , $GF ∥ A B$ , so $△ C GF \sim △ C A B$ (AA) and $\frac{GF}{A B} = \frac{C G}{C A} = \frac{4}{7}$ , giving $GF = \frac{40}{7}$ cm [1 mark].
$E F = \frac{72}{7} + \frac{40}{7} = \frac{112}{7} = 16$ cm [1 mark].
(iii) Let $\frac{A E}{A D} = t$ . Then $\frac{A G}{A C} = t$ , so $E G = 24 t$ and $GF = 10 (1 - t)$ , giving $E F = 10 + 14 t$ . $10 + 14 t = 17$ gives $t = \frac{1}{2}$ , so E must be the midpoint of AD [1 mark]. Check with (ii): $t = \frac{3}{7}$ gives $10 + 6 = 16$ cm.
Q7 · 5 marks · Long answerHarder board level
In $△ A B C$ , $∠ A B C = 9 0^{\circ}$ and BD is perpendicular to AC, with D on AC. $A D = 9$ cm and $D C = 16$ cm.
(i) Prove that $△ A D B \sim △ B D C$ .
(ii) Hence prove that $B D^{2} = A D \cdot D C$ and find BD.
(iii) Find AB and BC, using Pythagoras as a known result.
Hint
Show $∠ B A D = ∠ D B C$ by noting both equal $9 0^{\circ} - ∠ A B D$ .
Answer
(i) $∠ A D B = ∠ B D C = 9 0^{\circ}$ ; $∠ B A D = 9 0^{\circ} - ∠ A B D = ∠ D B C$ . So $△ A D B \sim △ B D C$ (AA) [2 marks].
(ii) $\frac{A D}{B D} = \frac{B D}{D C}$ , so $B D^{2} = A D \cdot D C = 144$ and $B D = 12$ cm [2 marks].
(iii) $A B = 81 + 144 = 15$ cm and $B C = 144 + 256 = 20$ cm. Check: $\frac{A B}{B C} = \frac{15}{20} = \frac{9}{12} = \frac{A D}{B D}$ [1 mark].
Q8 · 3 marks · Short answerHarder board level
In $△ A B C$ , $A B = 6$ cm, $B C = 8$ cm and $C A = 10$ cm. In $△ P QR$ , $P Q = 9$ cm, $QR = 12$ cm and $R P = 15$ cm. X is a point on BC with $B X = 2.5$ cm and $A X = 6.5$ cm. Y is a point on QR such that $∠ QP Y = ∠ B A X$ .
(i) Prove that $∠ A B C = ∠ P QR$ .
(ii) Prove that $△ A B X \sim △ P Q Y$ and find QY and PY.
Hint
You are not told any angle at B or Q. Get $∠ B = ∠ Q$ from a similarity of the whole triangles first.
Answer
(i) $\frac{A B}{P Q} = \frac{6}{9}$ , $\frac{B C}{QR} = \frac{8}{12}$ , $\frac{C A}{R P} = \frac{10}{15}$ , all equal to $\frac{2}{3}$ , so $△ A B C \sim △ P QR$ (SSS) and hence $∠ A B C = ∠ P QR$ [1 mark].
(ii) In $△ A B X$ and $△ P Q Y$ : $∠ A B X = ∠ P Q Y$ (from (i)) and $∠ B A X = ∠ QP Y$ (given), so $△ A B X \sim △ P Q Y$ (AA) [1 mark].
$\frac{A B}{P Q} = \frac{B X}{Q Y} = \frac{A X}{P Y} = \frac{2}{3}$ , so $Q Y = \frac{3}{2} \times 2.5 = 3.75$ cm and $P Y = \frac{3}{2} \times 6.5 = 9.75$ cm [1 mark].
Q9 · 3 marks · Case-basedBoard level
A pinhole camera is a closed box with a tiny hole in the front face and a screen at the back. A tree 2.4 m tall stands upright 6 m in front of the hole. The screen is parallel to the tree and 15 cm behind the hole, and an inverted image of the tree forms on it. Rays from the top and foot of the tree pass straight through the hole.
(i) Find the height of the image.
(ii) How far from the tree must the camera be placed for the image to be 9 cm tall?
Hint
The two triangles meet at the pinhole: use vertically opposite angles and alternate angles from the parallel tree and screen.
Answer
Triangles formed by the tree and the image are similar (AA: vertically opposite angles at the hole, alternate angles since tree is parallel to screen), so $\frac{image}{2.4} = \frac{0.15}{6}$ .
(i) Image $= 0.06$ m $= 6$ cm.
(ii) $\frac{0.09}{2.4} = \frac{0.15}{d}$ , so $d = \frac{2.4 \times 0.15}{0.09} = 4$ m.
Q10 · 3 marks · Short answerBoard level
In $△ A B C$ , $∠ A = 4 0^{\circ}$ and $∠ B = 7 5^{\circ}$ . In $△ P QR$ , $∠ P = 7 5^{\circ}$ and $∠ R = 6 5^{\circ}$ . Aman says the triangles are not similar because no angle of $△ A B C$ is given as $6 5^{\circ}$ . Kabir says $△ A B C \sim △ P QR$ .
(i) Who is right, if anyone? Write the correct similarity statement.
(ii) If $A B = 4$ cm, $P Q = 6$ cm and $B C = 5$ cm, find PR.
Hint
Find the third angle of each triangle using the angle sum of $18 0^{\circ}$ .
Answer
$∠ C = 18 0^{\circ} - 11 5^{\circ} = 6 5^{\circ}$ and $∠ Q = 18 0^{\circ} - 14 0^{\circ} = 4 0^{\circ}$ . So $A \leftrightarrow Q$ , $B \leftrightarrow P$ , $C \leftrightarrow R$ : $△ A B C \sim △ QP R$ (AA). Both are wrong: Aman missed the third angle and Kabir used the wrong order.
$\frac{B C}{P R} = \frac{A B}{QP} = \frac{4}{6}$ , so $P R = 7.5$ cm.
Q11 · 3 marks · Short answerHOTS / competency
In $△ A B C$ , $A B = 6$ cm, $B C = 9$ cm and $C A = 7.5$ cm. D is a point on BC such that $∠ B A D = ∠ B C A$ . Joining AD gives a figure with three triangles: ABC, ABD and ADC.
(i) Which two of these three triangles are similar? Prove it.
(ii) Prove that $A B^{2} = B C \cdot B D$ and find BD and DC.
(iii) Find AD. Show that $△ A D C$ is isosceles, and hence explain why $△ A D C$ is not similar to $△ A B C$ .
Hint
The given equal angle and a common angle must sit in the same two triangles. Which triangle contains both $∠ B A D$ and $∠ B$ ?
Answer
(i) In $△ A B D$ and $△ C B A$ : $∠ B A D = ∠ B C A$ (given) and $∠ A B D = ∠ C B A$ (common angle B). So $△ A B D \sim △ C B A$ (AA) [1 mark].
(ii) $\frac{A B}{C B} = \frac{B D}{B A}$ , so $A B^{2} = B C \cdot B D$ . Then $36 = 9 \times B D$ , so $B D = 4$ cm and $D C = 9 - 4 = 5$ cm [1 mark].
(iii) $\frac{A D}{C A} = \frac{A B}{C B} = \frac{6}{9} = \frac{2}{3}$ , so $A D = \frac{2}{3} \times 7.5 = 5$ cm. So $A D = D C = 5$ cm and $△ A D C$ is isosceles. $△ A B C$ has sides 6 cm, 7.5 cm and 9 cm, all different, so it has no two equal angles; an isosceles triangle cannot be similar to it [1 mark].
Q12 · 4 marks · Case-basedHOTS / competency
Suppose a town planner sketches a triangular park ABC. A straight fence DE runs parallel to the road BC, with D on AB and E on AC. On her sketch, AD and BC have the same length, $x$ m. She also notes $D B = 8$ m, $D E = 4$ m and $A C = 20$ m.
(i) Prove that $△ A D E \sim △ A B C$ and write the ratio of sides linking DE, BC, AD and AB. (1 mark)
(ii) Form an equation in $x$ . (1 mark)
(iii) Solve it to find $x$ , and then find AB and AE. (2 marks)
Hint
Use the whole side $A B = x + 8$ in the ratio for DE. The unknown appears twice, so expect a quadratic.
Answer
(i) $∠ A$ is common and $∠ A D E = ∠ A B C$ (corresponding angles, $D E ∥ B C$ ), so $△ A D E \sim △ A B C$ (AA) and $\frac{D E}{B C} = \frac{A D}{A B}$ [1 mark].
(ii) $A B = x + 8$ , so $\frac{4}{x} = \frac{x}{x + 8}$ . Cross-multiplying, $x^{2} = 4 x + 32$ , i.e. $x^{2} - 4 x - 32 = 0$ [1 mark].
(iii) $(x - 8) (x + 4) = 0$ , so $x = 8$ or $x = - 4$ . A length cannot be negative, so $x = 8$ m [1 mark].
$A B = 8 + 8 = 16$ m. $\frac{A E}{A C} = \frac{A D}{A B} = \frac{8}{16}$ , so $A E = \frac{1}{2} \times 20 = 10$ m. Check: $\frac{D E}{B C} = \frac{4}{8} = \frac{1}{2} = \frac{A D}{A B}$ [1 mark].

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the Basic Proportionality Theorem.Show answer
If a line is parallel to one side of a triangle and meets the other two sides in distinct points, it divides those sides in the same ratio: $\frac{A D}{D B} = \frac{A E}{E C}$ .
2.State the converse of BPT.Show answer
If a line divides two sides of a triangle in the same ratio, it is parallel to the third side.
3.Why are two equal angles enough for similarity?Show answer
The angle sum is $18 0^{\circ}$ , so the third angles are automatically equal (AA criterion).
4.What must be true about the angle in SAS similarity?Show answer
It must be the angle included between the two proportional sides.
5.Are all isosceles triangles similar?Show answer
No. Their angles can differ, for example $4 0^{\circ}, 7 0^{\circ}, 7 0^{\circ}$ and $10 0^{\circ}, 4 0^{\circ}, 4 0^{\circ}$ .
6.If $△ A B C \sim △ QR P$ , which side matches AB?Show answer
QR, the first two letters of the second triangle.
7.Ratio of perimeters of two similar triangles?Show answer
Equal to the ratio of any pair of corresponding sides.
8.In trapezium ABCD with $A B ∥ D C$ and diagonals meeting at O, what is $\frac{A O}{O C}$ ?Show answer
$\frac{A B}{C D}$ , because $△ A O B \sim △ C O D$ by AA.

Best order to study it

Start with similar figures: learn the two conditions (equal angles, proportional sides) and the standard counter examples (square and rhombus, square and rectangle).
Learn the meaning of $△ A B C \sim △ P QR$ and practise reading matching sides and angles from the letter order.
Learn the BPT statement and its proof with reasons, then practise numerical BPT questions including ones with $x$ .
Learn the converse of BPT and how it combines with AA to find the parallel side DE.
Learn AA, SSS and SAS with their conditions; practise correspondence with jumbled sides and proofs that end in a product of sides.
Finish with application problems: trapezium diagonals, shadows, lamp posts and case-based questions.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you write the full proof of BPT with construction and every reason?
Can you state the converse of BPT and say when to use it instead of BPT?
Can you find $x$ from a BPT equation and check the lengths are positive?
Can you find DE when $D E ∥ B C$ , using the whole side AB?
Can you state AA, SSS and SAS from memory, including the word "included" for SAS?
Can you write the correct similarity statement for two triangles with jumbled side lengths?
Can you give a counter example showing equal angles alone do not make two quadrilaterals similar?
Can you prove the diagonals of a trapezium divide each other in the ratio of the parallel sides?
Can you set up a lamp-post shadow problem with distance plus shadow as the base?
Can you find the perimeter of one similar triangle from the other's sides and one ratio?

Triangles: frequently asked questions

Is the areas of similar triangles theorem or Pythagoras theorem in the 2026-27 syllabus?

No. The theorem on areas of similar triangles and Pythagoras theorem with its converse as theorems of this chapter are not in the 2026-27 syllabus. You may still use Pythagoras as a known result where needed.

Which proof do I need to learn for this chapter?

Only the Basic Proportionality Theorem must be proved. Its converse and the AA, SSS and SAS criteria are stated without proof, but you must state them correctly and use them with reasons.

Is Thales theorem the same as the Basic Proportionality Theorem?

Yes. Both names refer to the same result: a line parallel to one side of a triangle divides the other two sides in the same ratio.

What is the difference between similar and congruent triangles?

Similar triangles have the same shape, with equal angles and proportional sides. Congruent triangles have the same shape and the same size, so the ratio of sides is exactly one. Every pair of congruent triangles is similar, but not the other way round.

How do I avoid mistakes with the order of letters?

Decide which vertex matches which before you write anything, using the equal angles or the sides that meet at each vertex. Then write both triangles in that order and read every ratio straight from the letter positions.

How should I write a similarity proof to get full marks?

Name the two triangles, list each pair of equal angles or proportional sides with a reason in brackets, and name the criterion at the end. Then write the ratio of corresponding sides in the correct order before deducing anything.

Based on: CBSE Mathematics syllabus 2026-27; NCERT Mathematics, Textbook for Class X (Reprint 2026-27)