MathsCBSE Class 10 · Triangles

Problems on Similar Triangles for Class 10

Updated 14 October 2026 · Fact-checked

Problems on similar triangles are mixed questions where you spot two similar triangles in a figure, prove them similar with AA, SSS or SAS, or use the Basic Proportionality Theorem, and then use the equal ratios of sides. The board asks you to find unknown lengths, prove products of sides equal, and solve real-life case studies.

Find the hidden similar pair, prove it by AA using equal angles at D and B, then write equal side ratios to get the unknown length.

Illustration drawn for QuestionBuddy.

Understand Problems on Similar Triangles

A problem on similar triangles almost never says "these triangles are similar". You must find the pair yourself, prove it, and then use it. All the chapter's tools come together here: the Basic Proportionality Theorem (BPT), its converse, and the AA, SSS and SAS criteria. Remember the definition: two triangles are similar when their corresponding angles are equal and their corresponding sides are in the same ratio. A counter example shows why both parts matter for other shapes: a square and a rhombus that is not a square, both of side 5 cm, have proportional sides but unequal angles, so they are not similar. (The proof of BPT itself is on the Basic Proportionality Theorem page.)

The reason similarity is so useful is simple. Once $△ A B C \sim △ P QR$ , every side of one is the same multiple $k$ of the matching side of the other. So a single known ratio unlocks every unknown length, and the perimeters are in the same ratio. Because $\frac{A B}{P Q} = \frac{A C}{P R}$ , you can also cross-multiply to get product results such as $A B \cdot P R = A C \cdot P Q$ . Most "prove that" questions are just this cross-multiplication in disguise.

How do you choose the tool? Look at what is given.

A line parallel to a side, and lengths on the other two sides: use BPT.
Lengths that may divide two sides in the same ratio, and you must show a line is parallel: use the converse of BPT.
Parallel lines, perpendiculars, a common angle or vertically opposite angles: get two equal angles and use AA.
All six sides known: use SSS.
One equal angle with the two sides forming it known: use SAS.

For a product result like $A C^{2} = B C \cdot D C$ , work backwards. Write it as a ratio, $\frac{A C}{B C} = \frac{D C}{A C}$ . The top letters ( $A, C, D$ ) name one triangle and the bottom letters ( $B, C, A$ ) name the other. Now you know which two triangles to prove similar.

The conditions matter. AA needs two pairs of equal angles with reasons. SAS needs the equal angle to lie between the two proportional sides; SSA is not a criterion. And the order of letters in a similarity statement fixes which sides match. Areas of similar triangles are not in this syllabus, so every problem here uses only lengths and angles. Pythagoras may be used as a known result when a right angle appears.

Problems on Similar Triangles: formulas to remember

Ratios from similarity: $△ A B C \sim △ P QR \Rightarrow \frac{A B}{P Q} = \frac{B C}{QR} = \frac{C A}{R P} = k$; Read matching sides from the letter order. The perimeters are in the same ratio k.
Basic Proportionality Theorem: $D E ∥ B C \Rightarrow \frac{A D}{D B} = \frac{A E}{E C}$; D on AB, E on AC. Part to part on both sides.
Part-to-whole form and third side: $D E ∥ B C \Rightarrow \frac{A D}{A B} = \frac{A E}{A C} = \frac{D E}{B C}$; The DE ratio comes from $△ A D E \sim △ A B C$ (AA). Always use the whole side AB with DE, never DB.
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.
SSS and SAS criteria: $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{C A}{R P} \Rightarrow △ A B C \sim △ P QR (SSS); ∠ A = ∠ P, \frac{A B}{P Q} = \frac{A C}{P R} \Rightarrow △ A B C \sim △ P QR (SAS)$; Stated without proof. SSS needs all three ratios equal. In SAS the equal angle must be included between the two proportional sides.
Common angle product result: $D on B C, ∠ A D C = ∠ B A C \Rightarrow △ A D C \sim △ B A C \Rightarrow C A^{2} = C B \cdot C D$; Angle C is common, so AA applies. Prove the similarity before quoting the result.
Altitude to the hypotenuse: $∠ B A C = 9 0^{\circ}, A D ⊥ B C \Rightarrow △ D B A \sim △ D A C \Rightarrow A D^{2} = B D \cdot D C$; Each smaller triangle is similar to the other by AA. Prove it in the exam each time.
Two altitudes of a triangle: $A D ⊥ B C, B E ⊥ A C \Rightarrow △ A D C \sim △ B E C \Rightarrow A C \cdot E C = B C \cdot D C$; Common angle C and two right angles give AA.

How to solve Problems on Similar Triangles questions

Use this routine on any similar triangles problem, from a 2-mark length question to a 5-mark proof.

1Read the figure description and mark every given length, parallel line, perpendicular and equal angle on a rough sketch.
2Decide what you need: a length, a parallel line, or a product or ratio of sides.
3For a product like $P A \cdot P B = P C \cdot P D$ , rewrite it as $\frac{P A}{P C} = \frac{P D}{P B}$ ; the top letters and bottom letters name the two triangles to use.
4Choose the tool: BPT or its converse for a line cutting two sides, AA for angles from parallels, perpendiculars or common angles, SSS for six sides, SAS for an included angle with two sides.
5Write each equal angle or ratio with its reason (common, vertically opposite, alternate, corresponding, $9 0^{\circ}$ ) and name the criterion.
6Write the similarity statement with vertices in matching order, then the ratio of sides from that order.
7Substitute, cross-multiply and solve. Check that every length is positive and give units.

Problems on Similar Triangles in real life

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.

Problems on Similar Triangles

Imagine your school has a triangular garden plot ABC. The gardener wants a straight walking path DE from point D on side AB to point E on side AC, running parallel to side BC. He measures AD = 12 m, DB = 8 m, AC = 30 m and BC = 25 m. Suppose edging for the path costs Rs 40 per metre.

Find AE and EC. Then find the length of the path DE and the cost of edging it along one side.

Try it first, then show the solutionSolution

Since $D E ∥ B C$ , by BPT (part-to-whole form): $\frac{A D}{A B} = \frac{A E}{A C}$ .
$A B = A D + D B = 12 + 8 = 20$ m.
$\frac{12}{20} = \frac{A E}{30}$ , so $A E = 18$ m.
$E C = A C - A E = 30 - 18 = 12$ m.
In $△ A D E$ and $△ A B C$ : $∠ A$ is common and $∠ A D E = ∠ A B C$ (corresponding angles, $D E ∥ B C$ ), so $△ A D E \sim △ A B C$ by AA.
$\frac{D E}{B C} = \frac{A D}{A B}$ , so $\frac{D E}{25} = \frac{12}{20}$ , giving $D E = 15$ m.
Cost $= 15 \times 40 =$ Rs 600.

Answer: AE = 18 m, EC = 12 m, path DE = 15 m, edging cost Rs 600.

Memory hook: A path parallel to the fence cuts a smaller copy of the plot: same shape, scaled by AD over AB.

Tricks to remember Problems on Similar Triangles

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

Top letters, bottom letters
Turn a product into a ratio. In $\frac{A C}{B C} = \frac{D C}{A C}$ the tops A, C, D form one triangle and the bottoms B, C, A form the other.
Use it when: Use it to pick the two triangles in any "prove that product equals product" question.
PAC-VR for equal angles
Parallel (alternate or corresponding), Angle common, Corner of $9 0^{\circ}$ , Vertically opposite, Remaining angle by angle sum.
Use it when: Run through this list when you need two equal angles for AA.
Whole with DE
DE pairs with the whole: $\frac{D E}{B C} = \frac{A D}{A B}$ . Parts pair with parts only in BPT.
Use it when: Stops the most common length error when finding the parallel segment.

Problems on Similar Triangles: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Finding the third side and a part using a parallel line · 2 marks · Very short answerBoard level

In $△ A B C$ , P is a point on AB and Q is a point on AC such that $P Q ∥ B C$ . $A P = 3$ cm, $P B = 6$ cm, $P Q = 4$ cm and $A C = 7.5$ cm. Find BC and QC.

Try it first, then show the solutionSolution

In $△ A P Q$ and $△ A B C$ : $∠ A$ is common, $∠ A P Q = ∠ A B C$ (corresponding angles, $P Q ∥ B C$ ).
So $△ A P Q \sim △ A B C$ by AA. (½)
$A B = 3 + 6 = 9$ cm, and $\frac{P Q}{B C} = \frac{A P}{A B}$ gives $\frac{4}{B C} = \frac{3}{9}$ , so $B C = 12$ cm. (½)
$\frac{A Q}{A C} = \frac{A P}{A B}$ gives $\frac{A Q}{7.5} = \frac{1}{3}$ , so $A Q = 2.5$ cm. (½)
$QC = 7.5 - 2.5 = 5$ cm. (½)

Answer: BC = 12 cm and QC = 5 cm.

Example 2: Common angle gives a square result · 3 marks · Short answerBoard level

In $△ A B C$ , D is a point on side BC such that $∠ A D C = ∠ B A C$ . Prove that $C A^{2} = C B \cdot C D$ . If $C B = 16$ cm and $C D = 4$ cm, find CA.

Try it first, then show the solutionSolution

Rewrite the target: $\frac{C A}{C B} = \frac{C D}{C A}$ , so compare $△ A D C$ and $△ B A C$ .
In $△ A D C$ and $△ B A C$ : $∠ A D C = ∠ B A C$ (given). (½)
$∠ A C D = ∠ B C A$ (common angle C). (½)
So $△ A D C \sim △ B A C$ by AA. (½)
Matching sides in order: $\frac{A C}{B C} = \frac{D C}{A C}$ . (½)
Cross-multiplying, $C A^{2} = C B \cdot C D$ . (½)
$C A^{2} = 16 \times 4 = 64$ , so $C A = 8$ cm. (½)

Answer: Proved $C A^{2} = C B \cdot C D$ ; CA = 8 cm.

Example 3: Median bisects a parallel segment · 3 marks · Short answerHarder board level

In $△ A B C$ , D is on AB and E is on AC with $D E ∥ B C$ . The median AM (M is the midpoint of BC) meets DE at N. Prove that N is the midpoint of DE.

Try it first, then show the solutionSolution

In $△ A D N$ and $△ A B M$ : $∠ D A N = ∠ B A M$ (common), $∠ A D N = ∠ A B M$ (corresponding, $D E ∥ B C$ ). So $△ A D N \sim △ A B M$ (AA). (1)
Hence $\frac{D N}{B M} = \frac{A N}{A M}$ . (½)
Similarly $△ A N E \sim △ A M C$ (common angle, $∠ A E N = ∠ A C M$ corresponding), so $\frac{N E}{M C} = \frac{A N}{A M}$ . (1)
So $\frac{D N}{B M} = \frac{N E}{M C}$ ; since $B M = M C$ , $D N = N E$ , and N is the midpoint of DE. (½)

Answer: DN = NE, so N is the midpoint of DE.

Example 4: Two altitudes: similarity, products and lengths · 5 marks · Long answerHarder board level

In $△ A B C$ , AD is perpendicular to BC (D on BC) and BE is perpendicular to AC (E on AC). AD and BE meet at H.
(i) Prove that $△ A D C \sim △ B E C$ and hence $A C \cdot E C = B C \cdot D C$ .
(ii) If $B C = 12$ cm, $D C = 4$ cm and $A C = 8$ cm, find EC and the ratio $A D : B E$ .
(iii) Prove that $A H \cdot H D = B H \cdot H E$ .

Try it first, then show the solutionSolution

(i) In $△ A D C$ and $△ B E C$ : $∠ A D C = ∠ B E C = 9 0^{\circ}$ (given perpendiculars). (½)
$∠ A C D = ∠ B C E$ (common angle C), so $△ A D C \sim △ B E C$ by AA. (½)
So $\frac{A C}{B C} = \frac{D C}{E C} = \frac{A D}{B E}$ , and cross-multiplying the first two gives $A C \cdot E C = B C \cdot D C$ . (1)
(ii) $8 \cdot E C = 12 \cdot 4 = 48$ , so $E C = 6$ cm. (½)
$\frac{A D}{B E} = \frac{A C}{B C} = \frac{8}{12} = \frac{2}{3}$ , so $A D : B E = 2 : 3$ . (½)
(iii) In $△ A H E$ and $△ B H D$ : $∠ A E H = ∠ B D H = 9 0^{\circ}$ , $∠ A H E = ∠ B H D$ (vertically opposite). So $△ A H E \sim △ B H D$ by AA. (1)
Hence $\frac{A H}{B H} = \frac{H E}{H D}$ , giving $A H \cdot H D = B H \cdot H E$ . (1)

Answer: (i) Proved. (ii) EC = 6 cm, AD : BE = 2 : 3. (iii) Proved.

Example 5: Case study: a pinhole camera · 4 marks · Case-basedHOTS / competency

Suppose your science class builds a pinhole camera: a closed box with a tiny hole O in the front face and a screen on the back face, 20 cm behind the hole. A candle AB, 15 cm tall, stands upright 60 cm in front of the hole. Light from the top A passes through O to point A' on the screen, and light from the base B passes through O to B'. The candle and the screen are both vertical, so $A B ∥ A^{'} B^{'}$ , and B, O, B' lie on a horizontal line with distances OB = 60 cm and OB' = 20 cm.
(i) Which criterion shows $△ O A B \sim △ O A^{'} B^{'}$ ? Give the two pairs of equal angles. (1 mark)
(ii) Find the height of the image A'B'. (1 mark)
(iii) How far, and in which direction, must the candle be moved along the line BO so that the image becomes 6 cm tall? (2 marks)

Try it first, then show the solutionSolution

(i) $∠ A O B = ∠ A^{'} O B^{'}$ (vertically opposite) and $∠ A B O = ∠ A^{'} B^{'} O = 9 0^{\circ}$ (both vertical, on a horizontal line). So AA similarity. (1)
(ii) $\frac{A ^{'} B ^{'}}{A B} = \frac{O B ^{'}}{O B}$ , so $\frac{A ^{'} B ^{'}}{15} = \frac{20}{60}$ , giving $A^{'} B^{'} = 5$ cm. (1)
(iii) Let the new distance be $d$ . $\frac{6}{15} = \frac{20}{d}$ . (1)
$d = \frac{15 \times 20}{6} = 50$ cm, so the candle moves $60 - 50 = 10$ cm towards the hole. (1)

Answer: (i) AA. (ii) 5 cm. (iii) Move the candle 10 cm closer to the hole (to 50 cm).

Example 6: SSS similarity, a missing angle and perimeters · 3 marks · Short answerBoard level

In $△ A B C$ , $A B = 6$ cm, $B C = 8$ cm, $C A = 10$ cm and $∠ B = 9 0^{\circ}$ . In $△ P QR$ , $P Q = 9$ cm, $QR = 12$ cm and $R P = 15$ cm. Prove that $△ A B C \sim △ P QR$ . Hence find $∠ Q$ , the value of $∠ P + ∠ R$ , and the ratio of the perimeters of the two triangles.

Try it first, then show the solutionSolution

$\frac{A B}{P Q} = \frac{6}{9} = \frac{2}{3}$ and $\frac{B C}{QR} = \frac{8}{12} = \frac{2}{3}$ . (½)
$\frac{C A}{R P} = \frac{10}{15} = \frac{2}{3}$ . (½)
All three pairs of corresponding sides are proportional, so $△ A B C \sim △ P QR$ by SSS. (½)
Corresponding angles of similar triangles are equal and B matches Q, so $∠ Q = ∠ B = 9 0^{\circ}$ . (Check: $6^{2} + 8^{2} = 1 0^{2}$ , so the right angle at B is consistent with the sides.) (½)
By the angle sum property in $△ P QR$ , $∠ P + ∠ R = 18 0^{\circ} - 9 0^{\circ} = 9 0^{\circ}$ . (½)
Perimeters: $6 + 8 + 10 = 24$ cm and $9 + 12 + 15 = 36$ cm, so the ratio is $24 : 36 = 2 : 3$ , the same as the side ratio. (½)

Answer: $△ A B C \sim △ P QR$ (SSS); $∠ Q = 9 0^{\circ}$ ; $∠ P + ∠ R = 9 0^{\circ}$ ; ratio of perimeters 2 : 3.

Example 7: Converse of BPT, then a parallelogram · 3 marks · Short answerHarder board level

In $△ A B C$ , D is a point on AB and E is a point on AC with $A D = 4$ cm, $D B = 6$ cm, $A E = 6$ cm, $E C = 9$ cm and $B C = 15$ cm. A line through E parallel to AB meets BC at F.
(i) Show that $D E ∥ B C$ .
(ii) Find DE.
(iii) Find BF and FC.

Try it first, then show the solutionSolution

(i) $\frac{A D}{D B} = \frac{4}{6} = \frac{2}{3}$ and $\frac{A E}{E C} = \frac{6}{9} = \frac{2}{3}$ . (½)
The two ratios are equal, so by the converse of BPT, $D E ∥ B C$ . (½)
(ii) In $△ A D E$ and $△ A B C$ : $∠ A$ common, $∠ A D E = ∠ A B C$ (corresponding, $D E ∥ B C$ ), so $△ A D E \sim △ A B C$ by AA. (½)
$A B = 4 + 6 = 10$ cm, $\frac{D E}{B C} = \frac{A D}{A B}$ gives $\frac{D E}{15} = \frac{4}{10}$ , so $D E = 6$ cm. (½)
(iii) $D E ∥ B F$ and $E F ∥ D B$ , so BDEF is a parallelogram and $B F = D E = 6$ cm (opposite sides). (½)
$F C = 15 - 6 = 9$ cm. Check by BPT in $△ C A B$ with $E F ∥ A B$ : $\frac{C F}{F B} = \frac{C E}{E A} = \frac{9}{6}$ , and $\frac{9}{6} = \frac{3}{2}$ agrees. (½)

Answer: (i) $D E ∥ B C$ by the converse of BPT. (ii) DE = 6 cm. (iii) BF = 6 cm, FC = 9 cm.

Example 8: Two poles and crossing wires: two similarities · 5 marks · Long answerHarder board level

Two vertical poles AB and CD stand on level ground with feet B and D, where $A B = 6$ m and $C D = 3$ m. A wire joins the top A of the first pole to the foot D of the second, and another wire joins the top C of the second pole to the foot B of the first. The wires AD and BC cross at E. EF is drawn perpendicular to the ground, with F on BD.
(i) Prove that $\frac{1}{A B} + \frac{1}{C D} = \frac{1}{E F}$ .
(ii) Find the height EF.
(iii) Find the ratio $B F : F D$ .

Try it first, then show the solutionSolution

(i) In $△ D F E$ and $△ D B A$ : $∠ D F E = ∠ D B A = 9 0^{\circ}$ (both vertical on level ground), $∠ D$ common, so $△ D F E \sim △ D B A$ by AA. (1)
Hence $\frac{E F}{A B} = \frac{D F}{D B}$ . (½)
In $△ B F E$ and $△ B D C$ : $∠ B F E = ∠ B D C = 9 0^{\circ}$ , $∠ B$ common, so $△ B F E \sim △ B D C$ by AA. (1)
Hence $\frac{E F}{C D} = \frac{B F}{B D}$ . (½)
Adding: $\frac{E F}{A B} + \frac{E F}{C D} = \frac{D F + B F}{B D} = \frac{B D}{B D} = 1$ ; dividing by EF gives $\frac{1}{A B} + \frac{1}{C D} = \frac{1}{E F}$ . (1)
(ii) $\frac{1}{E F} = \frac{1}{6} + \frac{1}{3} = \frac{1}{2}$ , so $E F = 2$ m. (½)
(iii) $\frac{B F}{B D} = \frac{E F}{C D} = \frac{2}{3}$ , so $F D = \frac{1}{3} B D$ and $B F : F D = 2 : 1$ . Check: $\frac{D F}{D B} = \frac{E F}{A B} = \frac{2}{6} = \frac{1}{3}$ agrees. (½)

Answer: (i) Proved. (ii) EF = 2 m. (iii) BF : FD = 2 : 1.

Where students lose marks in Problems on Similar Triangles

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

Mistake: Using the part DB instead of the whole AB to find DE
BPT is about parts, but the ratio for DE comes from similar triangles ADE and ABC, whose sides are AD and AB.
Wrong working
$A D = 3$ , $D B = 6$ , $B C = 12$ : $\frac{D E}{12} = \frac{3}{6}$ , so $D E = 6$ cm.
Right working
$\frac{D E}{B C} = \frac{A D}{A B} = \frac{3}{9}$ , so $D E = 4$ cm.
Do this: Write the similarity $△ A D E \sim △ A B C$ first, then read sides from it.
Usually costs: The final answer mark and often the method mark.
Mistake: Writing the similarity in the wrong vertex order
The order decides which sides match, so a jumbled statement gives a wrong ratio.
Wrong working
For $∠ A D C = ∠ B A C$ : writes $△ A D C \sim △ A B C$ , then $\frac{A D}{A B} = \frac{D C}{B C}$ .
Right working
Match D with A (the equal angles) and C with C: $△ A D C \sim △ B A C$ , so $\frac{A C}{B C} = \frac{D C}{A C}$ .
Do this: Pair the equal angles first, then write both triangles in that order.
Usually costs: Most of the marks in a 3-mark proof.
Mistake: Stating similarity without reasons
Examiners award marks for each equal angle with its justification and the named criterion.
Wrong working
" $△ A H E \sim △ B H D$ , so $A H \cdot H D = B H \cdot H E$ ."
Right working
" $∠ A E H = ∠ B D H = 9 0^{\circ}$ (given); $∠ A H E = ∠ B H D$ (vertically opposite); so similar by AA."
Do this: Write each pair of equal parts with a bracketed reason, then the criterion.
Usually costs: 1 to 2 marks in a proof question.
Mistake: Using SSA as if it were SAS
SAS needs the equal angle between the two proportional sides.
Wrong working
$\frac{A B}{P Q} = \frac{A C}{P R}$ and $∠ B = ∠ Q$ , so similar by SAS.
Right working
The angle included by AB and AC is $∠ A$ , so you need $∠ A = ∠ P$ for SAS.
Do this: Check that the equal angle is formed by exactly the two sides in your ratios.
Usually costs: The whole proof is marked wrong.

Exam tips

In proof questions, write the two triangles, each equal angle with its reason, and the criterion on separate lines; this is where step marks sit.
For a product or square result, convert it to a ratio first to find the triangles, but present the proof starting from the similarity.
After finding x from BPT, substitute back and check that every length is positive.
In case studies, describe or sketch the figure with labels before writing ratios, and include units in every answer.
Areas of similar triangles are not in the syllabus, so answer only with lengths, ratios and angles.

Practice questions on Problems on Similar 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 $△ A B C$ , D on AB and E on AC are such that $D E ∥ B C$ and $\frac{A D}{D B} = \frac{2}{3}$ . If $B C = 15$ cm, then DE is:
(a) 10 cm
(b) 6 cm
(c) 9 cm
(d) 7.5 cm
Hint
Convert part-to-part into part-to-whole before using DE.
Answer
(b) 6 cm. $\frac{A D}{A B} = \frac{2}{5}$ , and $\frac{D E}{B C} = \frac{A D}{A B}$ gives $D E = \frac{2}{5} \times 15 = 6$ cm.
Q2 · 3 marks · Short answerBoard level
In $△ A B C$ , $∠ B A C = 9 0^{\circ}$ and AD is perpendicular to BC, with D on BC. Prove that $A D^{2} = B D \cdot D C$ . If $B D = 4$ cm and $D C = 9$ cm, find AD.
Hint
Compare $△ D B A$ and $△ D A C$ . Show $∠ B = ∠ D A C$ using angle sums.
Answer
$∠ A D B = ∠ C D A = 9 0^{\circ}$ . $∠ B + ∠ B A D = 9 0^{\circ}$ and $∠ D A C + ∠ B A D = 9 0^{\circ}$ , so $∠ B = ∠ D A C$ . Hence $△ D B A \sim △ D A C$ (AA), $\frac{B D}{A D} = \frac{A D}{D C}$ , so $A D^{2} = B D \cdot D C = 36$ and AD = 6 cm.
Q3 · 3 marks · Short answerHarder board level
ABCD is a parallelogram with AB = 9 cm and AD = 6 cm. Side AD is produced beyond D to a point E with DE = 3 cm. BE meets CD at F. Prove that $△ A B E \sim △ C F B$ , and find CF and DF.
Hint
Opposite angles of a parallelogram are equal, and AE is parallel to BC.
Answer
$∠ A = ∠ C$ (opposite angles of parallelogram), $∠ A E B = ∠ C B F$ (alternate angles, $A E ∥ B C$ ), so $△ A B E \sim △ C F B$ (AA). AE = 9 cm, CB = 6 cm. $\frac{A B}{C F} = \frac{A E}{C B}$ gives $\frac{9}{C F} = \frac{9}{6}$ , so CF = 6 cm and DF = 9 − 6 = 3 cm.
Q4 · 3 marks · Short answerHarder board level
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. Find $x$ , then find DE if $B C = 35$ cm.
Hint
Use BPT, cross-multiply; the $x^{2}$ terms cancel.
Answer
$(x + 1) (3 x + 4) = (x + 3) (3 x - 1)$ gives $3 x^{2} + 7 x + 4 = 3 x^{2} + 8 x - 3$ , so $x = 7$ . AD = 8, DB = 20, AB = 28 cm. $D E = 35 \times \frac{8}{28} = 10$ cm.
Q5 · 3 marks · Short answerBoard level
In $△ A B C$ , D is on AB and E is on AC with AD = 3 cm, DB = 6 cm, AE = 4 cm and AC = 12 cm. Aman writes: " $\frac{A D}{D B} = \frac{1}{2}$ and $\frac{A E}{A C} = \frac{1}{3}$ . These are unequal, so DE is not parallel to BC."
(i) Find his error.
(ii) Decide correctly whether $D E ∥ B C$ .
(iii) If BC = 15 cm, find DE.
Hint
Compare part with part or whole with whole, never mixed.
Answer
(i) He compared a part-to-part ratio with a part-to-whole ratio. (ii) EC = 8 cm, so $\frac{A E}{E C} = \frac{1}{2} = \frac{A D}{D B}$ ; by the converse of BPT, $D E ∥ B C$ . (iii) $△ A D E \sim △ A B C$ (AA), $D E = 15 \times \frac{3}{9} = 5$ cm.
Q6 · 1 mark · Assertion-reasonBoard level
Assertion (A): If in $△ A B C$ and $△ P QR$ , $\frac{A B}{P Q} = \frac{A C}{P R}$ and $∠ B = ∠ Q$ , then the two triangles must be similar.
Reason (R): If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, the two 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.
Hint
Which angle is included between sides AB and AC?
Answer
(d) A is false but R is true. R is the SAS criterion. In A the angle included by AB and AC is $∠ A$ , not $∠ B$ , so this is SSA, which does not guarantee similarity.
Q7 · 3 marks · Short answerHOTS / competency
In $△ A B C$ , $A B = 8$ cm, $A C = 6$ cm and $B C = 9$ cm. D is a point on AB with $A D = 3$ cm and E is a point on AC with $A E = 4$ cm.
(i) Prove that $△ A D E \sim △ A C B$ .
(ii) Find DE.
(iii) Explain why DE is not parallel to BC, even though the two triangles are similar.
Hint
Pair AD with AC and AE with AB. For (iii), test the BPT ratios.
Answer
(i) $\frac{A D}{A C} = \frac{3}{6} = \frac{1}{2}$ and $\frac{A E}{A B} = \frac{4}{8} = \frac{1}{2}$ , and $∠ D A E = ∠ C A B$ (common, included between these sides), so $△ A D E \sim △ A C B$ by SAS. (ii) $\frac{D E}{C B} = \frac{1}{2}$ , so $D E = 4.5$ cm. (iii) $\frac{A D}{D B} = \frac{3}{5}$ and $\frac{A E}{E C} = \frac{4}{2} = 2$ . These are unequal, but if DE were parallel to BC, BPT would make them equal. So DE is not parallel to BC. The similarity matches D with C (so $∠ A D E = ∠ A C B$ ), not D with B.
Q8 · 5 marks · Long answerHarder board level
ABCD is a trapezium with $A B ∥ D C$ . Its diagonals AC and BD meet at O.
(i) Prove that $\frac{O A}{O C} = \frac{O B}{O D}$ , and hence that $O A \cdot O D = O B \cdot O C$ .
(ii) If $A B = 2 D C$ , $A C = 18$ cm and $B D = 15$ cm, find OA, OC, OB and OD, and show that O divides each diagonal in the ratio 2 : 1.
(iii) A line through O parallel to AB meets AD at E. Find $A E : E D$ .
Hint
Alternate angles from $A B ∥ D C$ and vertically opposite angles at O give AA. For (iii), OE is also parallel to DC, so use BPT in $△ A D C$ .
Answer
(i) In $△ A O B$ and $△ C O D$ : $∠ O A B = ∠ O C D$ (alternate, $A B ∥ D C$ ), $∠ A O B = ∠ C O D$ (vertically opposite), so $△ A O B \sim △ C O D$ (AA). Hence $\frac{O A}{O C} = \frac{O B}{O D} = \frac{A B}{C D}$ , and cross-multiplying gives $O A \cdot O D = O B \cdot O C$ . (ii) $\frac{O A}{O C} = \frac{O B}{O D} = \frac{A B}{C D} = 2$ , so each diagonal is divided in the ratio 2 : 1. $O A = \frac{2}{3} \times 18 = 12$ cm, $O C = 6$ cm, $O B = \frac{2}{3} \times 15 = 10$ cm, $O D = 5$ cm. (iii) $O E ∥ A B ∥ D C$ , so in $△ A D C$ with E on AD and O on AC, BPT gives $\frac{A E}{E D} = \frac{A O}{O C} = 2$ , so $A E : E D = 2 : 1$ .
Q9 · 4 marks · Case-basedHOTS / competency
Case study: Imagine a 6 m tall lamp post LM stands on a flat footpath with its base at M. At night, Riya, who is 1.5 m tall, stands upright at point T on the footpath, 9 m from M. Her head is at S, and her shadow on the ground is TX, where X is the tip of the shadow. The points M, T, X lie on one straight line on the ground, and the lamp L, her head S and the tip X lie on one straight line.
(i) Name the criterion that shows $△ X T S \sim △ X M L$ , and give the two pairs of equal angles. (1 mark)
(ii) Find the length of her shadow. (1 mark)
(iii) How far, and in which direction, must Riya walk along the line MX so that her shadow becomes 4 m long? How far is the tip of the shadow from M then? (2 marks)
Hint
Riya and the lamp post are both vertical, and the angle at X is shared. Let the shadow be s; then XM = 9 + s.
Answer
(i) AA: $∠ X$ is common and $∠ X T S = ∠ X M L = 9 0^{\circ}$ (both vertical on level ground). (ii) $\frac{S T}{L M} = \frac{X T}{X M}$ , so $\frac{1.5}{6} = \frac{s}{9 + s}$ , giving $4 s = 9 + s$ and $s = 3$ m. (iii) With shadow 4 m at distance d from M: $\frac{1.5}{6} = \frac{4}{d + 4}$ , so $d + 4 = 16$ and $d = 12$ m. She must walk $12 - 9 = 3$ m further away from the lamp post, and the tip of the shadow is then 16 m from M.

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.Which criterion needs only two pairs of equal angles?Show answer
AA, because the third angles are then equal by the angle sum property.
2.If $D E ∥ B C$ in $△ A B C$ , which ratio equals $\frac{D E}{B C}$ ?Show answer
$\frac{A D}{A B}$ (or $\frac{A E}{A C}$ ), from $△ A D E \sim △ A B C$ .
3.What must be true of the angle in SAS similarity?Show answer
It must be included between the two proportional sides.
4.Two similar triangles have sides in ratio 2 : 3. What is the ratio of their perimeters?Show answer
2 : 3, the same as the sides.
5.In a right triangle with altitude AD to the hypotenuse BC, what is $A D^{2}$ ?Show answer
$B D \cdot D C$ .

Problems on Similar Triangles: frequently asked questions

How do I know which criterion to use in a problem?

Look at what is given. Parallel lines, perpendiculars or a shared angle usually lead to AA. If all six sides are given use SSS, and if one equal angle and the two sides around it are given use SAS.

Do I have to prove the similarity every time, even for standard results like AD squared equals BD times DC?

Yes. In the board exam these are not quoted theorems, so prove the similarity with reasons and then cross-multiply. Only BPT, its converse and the criteria may be quoted directly.

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

No, it was removed in the NCERT rationalisation and is not in the 2026-27 syllabus.

Can I use Pythagoras in similar triangle problems?

Yes, you may use it as a known result when a right angle appears. Its proof is not part of this chapter for 2026-27.

Are these questions like NCERT Exercise 6.2 and 6.3?

Exercise 6.2 has questions on BPT and its converse, and Exercise 6.3 has questions on the criteria and proofs of relations. The examples here are of those kinds, set at board level, so practising them prepares you for both.