MathsICSE Class 10 · Similarity

Similar Triangles: AA, SAS and SSS Conditions, Proofs and Solved Questions (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

Two triangles are similar when their corresponding angles are equal and corresponding sides are in the same ratio. ICSE uses three tests: AA (two pairs of equal angles), SAS (one equal angle between proportional sides) and SSS (all three sides proportional). Write the triangles with matching vertices in the same order before taking ratios.

Matching arcs show equal angles at A and P, B and Q, C and R; every side of PQR is k times its partner, so write the letters in matching order.

Illustration drawn for QuestionBuddy.

Understand Similar Triangles: AA, SAS and SSS

Similarity as a size change. Similar figures have the same shape but possibly different sizes: one is an enlargement or reduction of the other by a scale factor $k$ . Congruent figures are the special case $k = 1$ . The key word for similarity is proportionality.

Similar triangles. $△ A B C \sim △ P QR$ means $∠ A = ∠ P$ , $∠ B = ∠ Q$ , $∠ C = ∠ R$ and

$\frac{A B}{P Q} = \frac{B C}{QR} = \frac{C A}{R P} = k .$

The order of the letters tells you which parts correspond. The 2025 Q1(xi) report noted candidates who paired the wrong sides when the triangles were drawn in different orientations.

The three conditions (proofs not required by the CISCE syllabus):

AA: two angles of one triangle equal two angles of the other (the third pair is then equal too).
SAS: one angle equal and the sides including it proportional.
SSS: all three pairs of sides in the same ratio.

Where equal angles come from in proofs: a common angle; vertically opposite angles; alternate or corresponding angles with parallel lines; right angles; angles given equal. In 2024 Q6(iii) the right reason was 'vertically opposite angles', and the report notes candidates who wrote a 'common angle' instead. Name angles with three letters ( $∠ A P D$ ) so it is clear which angle you mean.

Basic proportionality theorem (BPT). If a line parallel to one side of a triangle cuts the other two sides, it divides them in the same ratio: with $D E ∥ B C$ ,

$\frac{A D}{D B} = \frac{A E}{E C}$ , and $△ A D E \sim △ A B C$ with $\frac{D E}{B C} = \frac{A D}{A B}$ .

Careful: $\frac{D E}{B C}$ equals $\frac{A D}{A B}$ , not $\frac{A D}{D B}$ . The converse also holds: if $\frac{A D}{D B} = \frac{A E}{E C}$ , then $D E ∥ B C$ .

Writing a similarity proof.

1. Name the two triangles in matching order.
2. Give two pairs of equal angles with reasons (or the SAS/SSS data).
3. Conclude ' $∴ △ \dots \sim △ \dots$ (AA)'.
4. Write the equal ratios of sides from the letter order, then substitute.

Similar Triangles: AA, SAS and SSS: formulas to remember

Similar triangles: $△ A B C \sim △ P QR \Rightarrow \frac{A B}{P Q} = \frac{B C}{QR} = \frac{C A}{R P}$; Ratios follow the letter order.
Conditions: $AA, SAS, SSS$; Proofs of the conditions are not required.
Basic proportionality theorem: $D E ∥ B C \Rightarrow \frac{A D}{D B} = \frac{A E}{E C}$; And the converse.
Parallel line: third side: $\frac{D E}{B C} = \frac{A D}{A B} = \frac{A E}{A C}$; Use whole sides, not parts.

How to solve Similar Triangles: AA, SAS and SSS questions

Find the equal angles, fix the vertex order, then write the ratios.

1Mark the given equal angles, right angles, parallel lines and vertically opposite angles on the figure.
2Choose the two triangles and match vertices with equal angles.
3State the condition (AA, SAS or SSS) with reasons for each pair.
4Write the three ratios in the letter order of your similarity statement.
5Substitute known lengths into a pair of ratios with one unknown and solve.
6If Pythagoras is needed for a missing side, find it first.

Similar Triangles: AA, SAS and SSS 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.

Similar triangles to measure a height

Suppose you place a small mirror flat on the ground 10 m from a tree and step back until you see the treetop in it. You are then 2 m from the mirror and your eyes are 1.5 m above the ground.

Find the height of the tree.

Try it first, then show the solutionSolution

The light reflects at equal angles, and you and the tree both stand vertically, so the two triangles are similar (AA).
$\frac{tree}{10} = \frac{1.5}{2}$ , so tree $= 7.5$ m.

Answer: 7.5 m

Memory hook: Equal angles, proportional sides: small triangle measures the big one.

Enlargement keeps the shape

Imagine a photo 10 cm wide and 15 cm high that is enlarged so that its width becomes 25 cm without distortion.

Find the new height.

Try it first, then show the solutionSolution

Scale factor $k = \frac{25}{10} = 2.5$ .
New height $= 15 \times 2.5 = 37.5$ cm.

Answer: 37.5 cm

Memory hook: One scale factor for every length.

Tricks to remember Similar Triangles: AA, SAS and SSS

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

Order is everything
In $△ A B C \sim △ P QR$ , read across: A↔P, B↔Q, C↔R; so $A B$ matches $P Q$ , never $QR$ .
Use it when: Writing correct ratios.
AA is usually enough
Look first for a common angle, vertically opposite angles, right angles or parallel lines.
Use it when: Starting any similarity proof.
Whole over whole for the third side
$\frac{D E}{B C} = \frac{A D}{A B}$ : small triangle over big triangle.
Use it when: BPT questions.

Similar Triangles: AA, SAS and SSS: solved examples

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

Example 1: A missing side · 1 mark · MCQBoard level

$△ A B C \sim △ D E F$ with $A B = 4$ cm, $B C = 6$ cm and $D E = 6$ cm. Then $E F$ is:
(a) 4 cm
(b) 8 cm
(c) 9 cm
(d) 12 cm

Try it first, then show the solutionSolution

$\frac{E F}{B C} = \frac{D E}{A B} = \frac{6}{4} = 1.5$ , so $E F = 9$ cm.

Answer: (c) 9 cm

Example 2: Testing SSS · 2 marks · Short answerBoard level

The sides of one triangle are 5 cm, 7 cm and 9 cm; those of another are 7.5 cm, 10.5 cm and 13.5 cm. Are the triangles similar? Give the scale factor.

Try it first, then show the solutionSolution

Pair smallest with smallest: $\frac{7.5}{5} = 1.5$ , $\frac{10.5}{7} = 1.5$ , $\frac{13.5}{9} = 1.5$ .
All three ratios are equal, so the triangles are similar (SSS).

Answer: Yes (SSS); scale factor 1.5.

Example 3: Basic proportionality theorem · 3 marks · Short answerHarder board level

In $△ A B C$ , $D E ∥ B C$ with $D$ on $A B$ and $E$ on $A C$ . $A D = 4$ cm, $D B = 6$ cm and $A E = 5$ cm. Find $E C$ and $D E : B C$ .

Try it first, then show the solutionSolution

BPT: $\frac{A D}{D B} = \frac{A E}{E C} \Rightarrow \frac{4}{6} = \frac{5}{E C} \Rightarrow E C = 7.5$ cm.
$△ A D E \sim △ A B C$ , so $\frac{D E}{B C} = \frac{A D}{A B} = \frac{4}{10} = \frac{2}{5}$ .

Answer: $E C = 7.5$ cm; $D E : B C = 2 : 5$

Example 4: Proof with a common angle and right angles · 4 marks · Short answerHarder board level

In $△ A B C$ , $∠ A C B = 90°$ , $A C = 12$ cm and $B C = 9$ cm. $D$ is a point on $A B$ and $E$ a point on $A C$ such that $D E ⊥ A C$ and $D E = 3$ cm.
(a) Prove that $△ A E D \sim △ A C B$ .
(b) Find $A B$ , $A E$ and $A D$ .

Try it first, then show the solutionSolution

(a) In $△ A E D$ and $△ A C B$ : $∠ E A D = ∠ C A B$ (common) and $∠ A E D = ∠ A C B = 90°$ . So $△ A E D \sim △ A C B$ (AA).
(b) $A B = 1 2^{2} + 9^{2} = 15$ cm.
$\frac{A E}{A C} = \frac{E D}{C B} = \frac{A D}{A B}$ : $\frac{A E}{12} = \frac{3}{9} = \frac{A D}{15}$ .
$A E = 4$ cm, $A D = 5$ cm.

Answer: (b) $A B = 15$ cm, $A E = 4$ cm, $A D = 5$ cm

Example 5: Medians of similar triangles · 4 marks · Short answerHOTS / competency

$△ A B C \sim △ P QR$ . $A D$ and $P S$ are medians. Prove that $\frac{A D}{P S} = \frac{A B}{P Q}$ . If $A B = 6$ cm, $P Q = 9$ cm and $A D = 5$ cm, find $P S$ .

Try it first, then show the solutionSolution

From the similarity, $∠ B = ∠ Q$ and $\frac{A B}{P Q} = \frac{B C}{QR}$ .
Since $B D = \frac{1}{2} B C$ and $QS = \frac{1}{2} QR$ : $\frac{B D}{QS} = \frac{B C}{QR} = \frac{A B}{P Q}$ .
So $△ A B D \sim △ P QS$ (SAS), giving $\frac{A D}{P S} = \frac{A B}{P Q}$ .
$P S = 5 \times \frac{9}{6} = 7.5$ cm.

Answer: Proved; $P S = 7.5$ cm.

Example 6: Case study: a street lamp and a shadow · 4 marks · Case-basedHOTS / competency

A girl 1.6 m tall stands 4.8 m from the foot of a street lamp. Her shadow on level ground is 2.4 m long.
(a) Explain why the triangle formed by the girl and her shadow is similar to the triangle formed by the lamp post and the tip of the shadow.
(b) Find the height of the lamp.

Try it first, then show the solutionSolution

(a) Both the girl and the post are vertical (right angles with the ground), and both triangles share the angle at the tip of the shadow: AA similarity.
(b) The big triangle's base is $4.8 + 2.4 = 7.2$ m.
$\frac{H}{1.6} = \frac{7.2}{2.4} = 3$ , so $H = 4.8$ m.

Answer: (b) 4.8 m

Where students lose marks in Similar Triangles: AA, SAS and SSS

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

Mistake: Wrong reason for equal angles
Each pair of equal angles needs the correct reason.
Wrong working
$∠ A P D = ∠ B P C$ (common angle).
Right working
$∠ A P D = ∠ B P C$ (vertically opposite angles).
Do this: Check whether the angle is really shared or just opposite.
Usually costs: Reported in 2024 Q6(iii)(a).
Mistake: Pairing the wrong sides
Corresponding sides join corresponding vertices.
Wrong working
$△ A B C \sim △ E F G$ , so $\frac{A B}{F G} = \frac{B C}{E F}$ .
Right working
$\frac{A B}{E F} = \frac{B C}{F G}$ .
Do this: Write the similarity statement first and copy letters from it.
Usually costs: Reported in 2025 Q1(xi) and 2025 Q2(iii).
Mistake: Using DB instead of AB for the third side
$\frac{D E}{B C}$ compares the small triangle with the whole triangle.
Wrong working
$\frac{D E}{B C} = \frac{A D}{D B} = \frac{4}{6}$
Right working
$\frac{D E}{B C} = \frac{A D}{A B} = \frac{4}{10}$
Do this: Say 'small triangle over big triangle'.
Usually costs: Accuracy marks.

Exam tips

2024 Q6(iii) (4 marks): $△ A P D \sim △ B P C$ by AA using right angles and vertically opposite angles; then $B D = 8$ cm by Pythagoras before using ratios.
2025 Q2(iii) (4 marks): common angle $A$ and a right angle gave the similarity; the report stresses naming angles with three letters.
When a part says 'Hence', use the result just found (the 2024 report notes this was ignored).
Give the condition used in brackets: (AA), (SAS) or (SSS).

Practice questions on Similar Triangles: AA, SAS and SSS

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
If $△ A B C \sim △ X Y Z$ , then $\frac{A B}{X Y}$ equals:
(a) $\frac{B C}{X Z}$
(b) $\frac{B C}{Y Z}$
(c) $\frac{A C}{Y Z}$
(d) $\frac{A B}{Y Z}$
Hint
B ↔ Y and C ↔ Z.
Answer
(b) $\frac{B C}{Y Z}$
Q2 · 2 marks · Short answerBoard level
In $△ A B C$ , $D E ∥ B C$ , $A D = 3$ cm, $A B = 9$ cm and $A E = 2$ cm. Find $A C$ .
Hint
$\frac{A D}{A B} = \frac{A E}{A C}$ .
Answer
$A C = 6$ cm
Q3 · 3 marks · Short answerHarder board level
In $△ A B C$ and $△ P QR$ , $∠ A = ∠ P = 50°$ , $A B = 4$ cm, $A C = 6$ cm, $P Q = 6$ cm and $P R = 9$ cm. Show that the triangles are similar. If $B C = 5$ cm, find $QR$ .
Hint
Compare $\frac{P Q}{A B}$ and $\frac{P R}{A C}$ .
Answer
$\frac{6}{4} = \frac{9}{6} = 1.5$ with the included angles equal: similar (SAS). $QR = 7.5$ cm.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Any two squares are similar.
Reason (R): Similar figures have equal corresponding angles and proportional corresponding sides.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
All angles are $90°$ and all sides in a square are equal.
Answer
(c)
Q5 · 2 marks · Short answerHarder board level
A 2 m stick casts a 3 m shadow at the same time as a tree casts a 15 m shadow on level ground. Find the height of the tree.
Hint
Parallel sun rays give similar triangles.
Answer
$\frac{h}{15} = \frac{2}{3}$ , so $h = 10$ m.
Q6 · 4 marks · Short answerHOTS / competency
In $△ A B C$ , $D$ is on $A B$ with $A D : D B = 1 : 2$ . $D E ∥ B C$ meets $A C$ at $E$ , and $E F ∥ A B$ meets $B C$ at $F$ . If $B C = 12$ cm, find $D E$ and $B F$ .
Hint
$B D E F$ is a parallelogram.
Answer
$\frac{D E}{B C} = \frac{A D}{A B} = \frac{1}{3}$ , so $D E = 4$ cm. $B D E F$ has both pairs of opposite sides parallel, so $B F = D E = 4$ cm.
Q7 · 4 marks · Short answerHarder board level
In $△ A B C$ , $∠ A B C = 90°$ , $A B = 6$ cm, $B C = 8$ cm and $B D ⊥ A C$ . Prove that $△ A D B \sim △ A B C$ and find $A D$ and $B D$ .
Hint
Common $∠ A$ and a right angle each.
Answer
$∠ A$ common, $∠ A D B = ∠ A B C = 90°$ : AA. $A C = 10$ ; $\frac{A D}{A B} = \frac{A B}{A C}$ gives $A D = 3.6$ cm; $\frac{B D}{B C} = \frac{A B}{A C}$ gives $B D = 4.8$ cm.

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.Name the three conditions for similar triangles.Show answer
AA, SAS and SSS.
2.State the basic proportionality theorem.Show answer
A line parallel to one side of a triangle divides the other two sides in the same ratio.
3.Are congruent triangles similar?Show answer
Yes, with scale factor 1.
4. $△ A B C \sim △ P QR$ and $A B = 2 P Q$ . What is $B C : QR$ ?Show answer
$2 : 1$
5.In SAS similarity, which angle must be equal?Show answer
The angle included between the two proportional sides.

Similar Triangles: AA, SAS and SSS: frequently asked questions

What are the conditions for two triangles to be similar?

AA (two pairs of equal angles), SAS (an equal angle included between proportional sides) and SSS (all three pairs of sides proportional). ICSE does not require proofs of these conditions.

What is the difference between similar and congruent triangles?

Congruent triangles have the same shape and size. Similar triangles have the same shape; their sides are proportional with some scale factor k, and congruence is the case k = 1.

How do I find a missing side using similar triangles?

Write the similarity with vertices in matching order, write the equal ratios of corresponding sides, and solve the ratio that contains the unknown.