MathsCBSE Class 10 · Triangles

SAS Similarity Criterion for Class 10 Triangles

Updated 14 October 2026 · Fact-checked

The SAS similarity criterion says two triangles are similar if one angle of one equals one angle of the other and the sides including those angles are in the same ratio. It is stated without proof. The board asks you to prove triangles similar with it, find missing sides and angles, and spot the SSA trap.

Check the equal angle sits between the two proportional sides (AB, AC and PQ, PR); then the triangles are similar.

Illustration drawn for QuestionBuddy.

Understand SAS Similarity Criterion

The SAS similarity criterion is a shortcut for proving two triangles similar. You check only three facts: one pair of equal angles, and the two pairs of sides that form those angles. If $∠ A = ∠ P$ and $\frac{A B}{P Q} = \frac{A C}{P R}$ , then $△ A B C \sim △ P QR$ . In the syllabus it is stated without proof, so you quote it and use it.

Why does it work? Two sides and the angle between them fix a triangle's shape completely. In Class 9 you saw that SAS fixes a triangle exactly, which is congruence. If you scale both sides by the same factor $k$ and keep the angle, the third side scales by $k$ too, and all the angles stay the same. So you get the same shape at a different size.

The key word is including. The equal angle must sit between the two sides you compare. In $△ A B C$ , the angle included between $A B$ and $A C$ is $∠ A$ , not $∠ B$ . If the equal angle is not between the proportional sides, you have an SSA situation. SSA is not a criterion, because two different shapes can share those facts.

SAS fits naturally where triangles share an angle: a common vertex angle, or vertically opposite angles where two segments cross. It also handles the reversed case, where a small triangle cut off at a vertex matches the big triangle with its sides swapped. Then you compare $\frac{A D}{A C}$ with $\frac{A E}{A B}$ . Test both pairings before you decide.

SAS sits beside the other criteria. Use AA when you have angles, SSS when you have all six sides, and SAS when you have one angle and two sides of each triangle. Once similarity is proved, you read off the third pair of sides and the other pairs of equal angles from the correspondence. Those equal angles often lead to parallel lines (equal alternate angles), and the side ratios lead to products such as $C A^{2} = C B \cdot C D$ .

SAS Similarity Criterion: formulas to remember

SAS similarity criterion: $∠ A = ∠ P and \frac{A B}{P Q} = \frac{A C}{P R} \Rightarrow △ A B C \sim △ P QR$; Stated without proof. The equal angle must be included between the two proportional sides.
What you get after SAS: $\frac{B C}{QR} = \frac{A B}{P Q}, ∠ B = ∠ Q, ∠ C = ∠ R$; Use the correspondence to find the third side and the other equal angles.
Reversed correspondence at a common vertex: $D \in A B, E \in A C, \frac{A D}{A C} = \frac{A E}{A B} \Rightarrow △ A D E \sim △ A C B$; Common angle A. Here $D E$ is not parallel to $B C$ in general, and $∠ A D E = ∠ A C B$ .
Crossing segments: $A D \cap B C = O, \frac{O A}{O D} = \frac{O B}{O C} \Rightarrow △ A O B \sim △ D O C \Rightarrow A B ∥ D C$; Vertically opposite angles are the included angle. Equal alternate angles then give the parallel lines.
Product form: $D \in B C, C A^{2} = C B \cdot C D \Rightarrow △ A D C \sim △ B A C$; Rewrite as $\frac{C A}{C B} = \frac{C D}{C A}$ with common angle C, then $∠ C A D = ∠ C B A$ .
Medians of similar triangles: $△ A B C \sim △ P QR, A D, P M medians \Rightarrow \frac{A D}{P M} = \frac{A B}{P Q}$; Proved by SAS on $△ A B D$ and $△ P QM$ , since $\frac{B D}{QM} = \frac{B C}{QR}$ .

How to solve SAS Similarity Criterion questions

Use this routine on any SAS question, whether it asks for a proof, a length or an angle.

1Find the equal angle and give its reason: given, common angle, or vertically opposite angles.
2Name the two sides that form that angle in each triangle. These are the only sides you may compare.
3Work out both ratios. If they do not match, test the swapped pairing (short side with long side) before concluding.
4When the two ratios are equal, write: by SAS similarity criterion, followed by the similarity statement with the vertices in matching order.
5Read the correspondence: equal angle to equal angle, and each side to the side between the matching vertices.
6Use the common ratio to find the third side, or use the equal angles to find an angle or prove lines parallel.
7Check the answer: the third-side ratio must equal the scale factor, and every length must be positive.

SAS Similarity Criterion 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.

SAS similarity criterion

Suppose your family has a folding clothes-drying stand. From the side it looks like two straight rods AD and BC crossing at a hinge O, with a top bar AB joining the upper ends. The lower ends D and C rest on the floor.

Along the rods, OA = 30 cm, OB = 36 cm, OD = 60 cm and OC = 72 cm. The top bar AB is 24 cm long. Prove that the triangles AOB and DOC are similar. Find the distance DC between the feet, and show that the top bar is parallel to the floor.

Try it first, then show the solutionSolution

In triangles AOB and DOC, angle AOB = angle DOC (vertically opposite angles).
OA/OD = 30/60 = 1/2 and OB/OC = 36/72 = 1/2, so OA/OD = OB/OC.
These sides include the equal angles, so triangle AOB ~ triangle DOC by SAS similarity.
AB/DC = 1/2, so DC = 2 × 24 = 48 cm.
From the similarity, angle OAB = angle ODC. These are alternate angles made by the transversal AD with AB and DC.
Equal alternate angles mean AB ∥ DC, so the top bar is parallel to the floor.

Answer: DC = 48 cm, and the top bar AB is parallel to the floor.

Memory hook: Crossed rods, a hinge angle and equal ratios: SAS makes the stand open evenly.

Tricks to remember SAS Similarity Criterion

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

The sandwich rule
The angle is the filling and the two sides are the bread. S-A-S means the angle sits between the sides. If it does not, the sandwich falls apart (SSA).
Use it when: Before writing SAS, check that the equal angle is formed by the two sides you compared.
Short with short, long with long
At the shared vertex, pair the shorter side of the small triangle with the shorter side of the big triangle, and the longer with the longer. If the ratios do not match, swap the pairing, then name the matching vertices.
Use it when: Handles the reversed case, where $△ A D E \sim △ A C B$ and not $△ A B C$ .
Square means SAS
When you see $C A^{2} = C B \cdot C D$ , read it as a ratio, $\frac{C A}{C B} = \frac{C D}{C A}$ . The repeated letter C gives the common angle.
Use it when: Product-form proofs in 3- and 5-mark questions.

SAS Similarity Criterion: solved examples

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

Example 1: Direct SAS and the third side · 2 marks · Very short answerBoard level

In $△ P QR$ and $△ X Y Z$ , $∠ P = ∠ X = 9 0^{\circ}$ , $P Q = 6$ cm, $P R = 8$ cm, $X Y = 3$ cm and $X Z = 4$ cm. Show that $△ P QR \sim △ X Y Z$ . If $Y Z = 5$ cm, find $QR$ .

Try it first, then show the solutionSolution

$\frac{P Q}{X Y} = \frac{6}{3} = 2$ and $\frac{P R}{X Z} = \frac{8}{4} = 2$ (½)
$∠ P = ∠ X$ , and it is included between these sides. (½)
So $△ P QR \sim △ X Y Z$ by SAS similarity. (½)
$\frac{QR}{Y Z} = 2$ , so $QR = 2 \times 5 = 10$ cm. Check: $6^{2} + 8^{2} = 100 = 1 0^{2}$ , as it must be for a right angle at P. (½)

Answer: The triangles are similar by SAS, and $QR = 10$ cm.

Example 2: Crossing segments: prove parallel · 3 marks · Short answerBoard level

Line segments AD and BC intersect at O. $O A = 4$ cm, $O D = 6$ cm, $O B = 6$ cm and $O C = 9$ cm. Prove that $△ A O B \sim △ D O C$ and that $A B ∥ D C$ . If $D C = 12$ cm, find AB.

Try it first, then show the solutionSolution

$∠ A O B = ∠ D O C$ (vertically opposite angles). (½)
$\frac{O A}{O D} = \frac{4}{6} = \frac{2}{3}$ and $\frac{O B}{O C} = \frac{6}{9} = \frac{2}{3}$ (½)
So $△ A O B \sim △ D O C$ by SAS similarity. (½)
Hence $∠ O A B = ∠ O D C$ . These are alternate angles made by the transversal AD, so $A B ∥ D C$ . (½)
$\frac{A B}{D C} = \frac{2}{3}$ (½)
$A B = \frac{2}{3} \times 12 = 8$ cm (½)

Answer: $△ A O B \sim △ D O C$ by SAS, $A B ∥ D C$ , and $A B = 8$ cm.

Example 3: Reversed correspondence at a common vertex · 3 marks · Short answerBoard level

In $△ A B C$ , $A B = 10$ cm, $A C = 8$ cm and $B C = 6$ cm. D is a point on AB with $A D = 4$ cm, and E is a point on AC with $A E = 5$ cm. Prove that $△ A D E \sim △ A C B$ . Find DE. Given that $∠ A C B = 9 0^{\circ}$ , find $∠ A D E$ .

Try it first, then show the solutionSolution

Same-order test: $\frac{A D}{A B} = \frac{4}{10}$ but $\frac{A E}{A C} = \frac{5}{8}$ . These are not equal, so swap the pairing. (½)
$\frac{A D}{A C} = \frac{4}{8} = \frac{1}{2}$ and $\frac{A E}{A B} = \frac{5}{10} = \frac{1}{2}$ (½)
$∠ D A E = ∠ C A B$ (common angle), included between these sides. (½)
So $△ A D E \sim △ A C B$ by SAS similarity. (½)
$\frac{D E}{C B} = \frac{1}{2}$ , so $D E = \frac{1}{2} \times 6 = 3$ cm. (½)
$∠ A D E = ∠ A C B = 9 0^{\circ}$ , since D corresponds to C. Check: $3^{2} + 4^{2} = 5^{2}$ , so triangle ADE is indeed right-angled at D. (½)

Answer: $△ A D E \sim △ A C B$ by SAS, $D E = 3$ cm and $∠ A D E = 9 0^{\circ}$ .

Example 4: Unknown in the ratio, then SAS, parallel lines and BPT · 5 marks · Long answerHarder board level

The diagonals AC and BD of a quadrilateral ABCD intersect at O. $O A = (x + 1)$ cm, $O C = (x + 7)$ cm, $O B = (x + 3)$ cm and $O D = (x + 12)$ cm.
(i) Find the value of $x$ for which $\frac{O A}{O C} = \frac{O B}{O D}$ .
(ii) For this value, prove that $△ A O B \sim △ C O D$ and that $A B ∥ D C$ .
(iii) If $A B = 8$ cm, find DC.
(iv) E is a point on AD such that $O E ∥ D C$ . Find the ratio $A E : E D$ .

Try it first, then show the solutionSolution

$\frac{x + 1}{x + 7} = \frac{x + 3}{x + 12}$ , so $(x + 1) (x + 12) = (x + 3) (x + 7)$ . (½)
$x^{2} + 13 x + 12 = x^{2} + 10 x + 21$ (½)
$3 x = 9$ , so $x = 3$ . Then $O A = 4$ cm, $O C = 10$ cm, $O B = 6$ cm, $O D = 15$ cm. (½)
$\frac{O A}{O C} = \frac{4}{10} = \frac{2}{5}$ , $\frac{O B}{O D} = \frac{6}{15} = \frac{2}{5}$ , and $∠ A O B = ∠ C O D$ (vertically opposite angles, included between these sides), so $△ A O B \sim △ C O D$ by SAS similarity. (½)
Hence $∠ O A B = ∠ O C D$ . These are alternate angles made by the transversal AC, so $A B ∥ D C$ . (1)
$\frac{A B}{C D} = \frac{O A}{O C} = \frac{2}{5}$ , so $C D = 8 \times \frac{5}{2} = 20$ cm. (1)
In $△ A D C$ , $O E ∥ D C$ , so by the Basic Proportionality Theorem $\frac{A E}{E D} = \frac{A O}{O C} = \frac{4}{10} = \frac{2}{5}$ , that is $A E : E D = 2 : 5$ . (1)

Answer: (i) $x = 3$ . (ii) Similar by SAS, and $A B ∥ D C$ . (iii) $D C = 20$ cm. (iv) $A E : E D = 2 : 5$ .

Example 5: Product form proof with numbers · 5 marks · Long answerHarder board level

In $△ A B C$ , D is a point on side BC such that $C A^{2} = C B \cdot C D$ .
(i) Prove that $△ A D C \sim △ B A C$ .
(ii) Hence prove that $∠ C A D = ∠ A B C$ .
(iii) If $C A = 6$ cm, $C B = 9$ cm and $A B = 7.5$ cm, find CD, BD and AD.

Try it first, then show the solutionSolution

Given $C A^{2} = C B \cdot C D$ , so $\frac{C A}{C B} = \frac{C D}{C A}$ . (1)
In $△ A D C$ and $△ B A C$ , $∠ A C D = ∠ B C A$ (common angle C). (½)
The sides including angle C are CA, CD in $△ A D C$ and CB, CA in $△ B A C$ , and $\frac{C A}{C B} = \frac{C D}{C A}$ . (½)
So $△ A D C \sim △ B A C$ by SAS similarity, with $A \leftrightarrow B$ , $D \leftrightarrow A$ , $C \leftrightarrow C$ . (1)
Corresponding angles are equal, so $∠ C A D = ∠ C B A = ∠ A B C$ . (½)
$C D = \frac{C A ^{2}}{C B} = \frac{36}{9} = 4$ cm, so $B D = 9 - 4 = 5$ cm. (½)
$\frac{A D}{B A} = \frac{A C}{B C} = \frac{6}{9} = \frac{2}{3}$ (½)
$A D = \frac{2}{3} \times 7.5 = 5$ cm (½)

Answer: (i) Similar by SAS. (ii) $∠ C A D = ∠ A B C$ . (iii) $C D = 4$ cm, $B D = 5$ cm, $A D = 5$ cm.

Example 6: Case study: width of a pond · 4 marks · Case-basedHOTS / competency

Imagine Arjun needs the width PQ of a pond in his village. He stands at a point O on dry land, where $O P = 40$ m and $O Q = 48$ m along straight lines to the two ends of the pond. He wants to mark A on OP and B on OQ so that $△ O A B \sim △ O P Q$ . He decides that OB will be 2 m longer than OA, so he takes $O A = x$ m and $O B = (x + 2)$ m. After marking, he measures $A B = 6.5$ m.
(i) Which two ratios must be equal for SAS similarity, and which angle must they include? (1)
(ii) Find $x$ , and hence OA and OB. (1)
(iii) Prove the similarity, find the width PQ, and show that AB is parallel to PQ. (2)

Try it first, then show the solutionSolution

(i) $\frac{O A}{O P} = \frac{O B}{O Q}$ . These sides lie along the two arms at O, so they include the common angle $∠ A O B = ∠ P O Q$ . (1)
(ii) $\frac{x}{40} = \frac{x + 2}{48}$ , so $48 x = 40 x + 80$ . (½)
$8 x = 80$ , so $x = 10$ . Then $O A = 10$ m and $O B = 12$ m. (½)
(iii) $\frac{O A}{O P} = \frac{10}{40} = \frac{1}{4}$ and $\frac{O B}{O Q} = \frac{12}{48} = \frac{1}{4}$ , with common $∠ O$ , so $△ O A B \sim △ O P Q$ by SAS similarity. (½)
$\frac{A B}{P Q} = \frac{1}{4}$ , so $P Q = 4 \times 6.5 = 26$ m. (½)
From the similarity, $∠ O A B = ∠ O P Q$ . These are corresponding angles made by the transversal OP, so $A B ∥ P Q$ . (1)

Answer: (i) $\frac{O A}{O P} = \frac{O B}{O Q}$ , including the common angle at O. (ii) $x = 10$ , so $O A = 10$ m and $O B = 12$ m. (iii) $P Q = 26$ m, and $A B ∥ P Q$ .

Where students lose marks in SAS Similarity Criterion

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

Mistake: Using an angle that is not included (SSA)
Students see one equal angle and two equal ratios and assume SAS applies.
Wrong working
$\frac{A B}{P Q} = \frac{B C}{QR}$ and $∠ A = ∠ P$ , so $△ A B C \sim △ P QR$ by SAS.
Right working
The sides AB and BC include $∠ B$ . You need $∠ B = ∠ Q$ , or else a different criterion.
Do this: Mark the angle on your rough figure and check that both compared sides touch it.
Usually costs: The whole proof mark, and every mark that depends on the similarity.
Mistake: Forcing the same-order pairing in the reversed case
Students assume D matches B because D lies on AB.
Wrong working
$\frac{A D}{A B} = \frac{4}{10} \neq = \frac{5}{8} = \frac{A E}{A C}$ , so the triangles are not similar.
Right working
$\frac{A D}{A C} = \frac{A E}{A B} = \frac{1}{2}$ with common angle A, so $△ A D E \sim △ A C B$ .
Do this: If one pairing fails, test the swapped pairing before concluding.
Usually costs: The full question, because the wrong conclusion is reached.
Mistake: Wrong vertex order in the statement
The letters are written alphabetically instead of by correspondence.
Wrong working
$△ A D E \sim △ A B C$ , so $∠ A D E = ∠ A B C$ .
Right working
$△ A D E \sim △ A C B$ , so $∠ A D E = ∠ A C B$ .
Do this: Write the matching vertices in pairs first ( $A \leftrightarrow A$ , $D \leftrightarrow C$ , $E \leftrightarrow B$ ), then write the statement.
Usually costs: ½ to 1 mark, plus any wrong angle or length that follows from it.
Mistake: No reason for the equal angle
Students treat the common or vertically opposite angle as obvious.
Wrong working
$∠ A O B = ∠ D O C$ , so the triangles are similar by SAS.
Right working
$∠ A O B = ∠ D O C$ (vertically opposite angles), then state the ratios, then write: by SAS similarity criterion.
Do this: Give every equality a reason in brackets.
Usually costs: ½ mark per missing reason.

Exam tips

Always name the criterion in words, for example: by SAS similarity criterion. Examiners award a step mark for it.
Write both ratios as fractions and simplify them to the same value. Do not just say the sides are proportional.
In assertion-reason and find-the-error items, check first whether the angle is included. This is a common slip.
After proving similarity, the question usually wants something more: a third side, an angle, a parallel line or a product. Read every part before you start.
Do not use the ratio of areas of similar triangles. That result is not in your syllabus.

Practice questions on SAS Similarity Criterion

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 $△ D E F$ , $∠ B = ∠ E$ , $A B = 4$ cm, $B C = 6$ cm and $D E = 6$ cm. For the two triangles to be similar by SAS, the length EF must be:
(a) 3 cm
(b) 8 cm
(c) 9 cm
(d) 12 cm
Hint
The sides including $∠ B$ are AB and BC. The sides including $∠ E$ are DE and EF.
Answer
(c) 9 cm. $\frac{A B}{D E} = \frac{B C}{E F}$ gives $\frac{4}{6} = \frac{6}{E F}$ , so $E F = 9$ cm. None of the other options satisfies either pairing.
Q2 · 1 mark · Assertion-reasonBoard level
Assertion (A): In $△ P QR$ and $△ X Y Z$ , $P Q = 3$ cm, $QR = 4$ cm, $X Y = 6$ cm, $Y Z = 8$ cm and $∠ R = ∠ Z$ . So $△ P QR \sim △ X Y Z$ by the SAS similarity criterion.
Reason (R): For SAS similarity, the equal angles must be included between the proportional sides.
(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 formed by sides PQ and QR?
Answer
(d) A is false but R is true. The ratios are both $\frac{1}{2}$ , but the sides PQ and QR include $∠ Q$ , not $∠ R$ . This is SSA, so SAS cannot be used. R correctly states the included-angle condition.
Q3 · 3 marks · Short answerHarder board level
$△ A B C \sim △ P QR$ . AD and PM are medians to BC and QR. Prove that $\frac{A D}{P M} = \frac{A B}{P Q}$ . If $A B = 8$ cm, $P Q = 6$ cm and $A D = 6.4$ cm, find PM.
Hint
Halve BC and QR, then apply SAS to $△ A B D$ and $△ P QM$ with $∠ B = ∠ Q$ .
Answer
From the similarity, $∠ B = ∠ Q$ and $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{2 B D}{2 QM} = \frac{B D}{QM}$ . So $△ A B D \sim △ P QM$ by SAS, which gives $\frac{A D}{P M} = \frac{A B}{P Q}$ . Then $P M = 6.4 \times \frac{6}{8} = 4.8$ cm.
Q4 · 4 marks · Long answerHarder board level
In $△ A B C$ , AD is perpendicular to BC, with D between B and C, and $A D^{2} = B D \times D C$ .
(i) Prove that $△ A D B \sim △ C D A$ .
(ii) Hence prove that $∠ B A C = 9 0^{\circ}$ .
(iii) If $B D = 4$ cm and $D C = 9$ cm, find AD, AB and AC, and check your answer with Pythagoras' theorem in $△ A B C$ .
Hint
Rewrite $A D^{2} = B D \times D C$ as a proportion of the sides that include the right angles at D.
Answer
(i) $A D^{2} = B D \times D C$ gives $\frac{A D}{C D} = \frac{B D}{A D}$ , and $∠ A D B = ∠ C D A = 9 0^{\circ}$ is included between these sides, so $△ A D B \sim △ C D A$ by SAS (with $A \leftrightarrow C$ , $D \leftrightarrow D$ , $B \leftrightarrow A$ ). (1)
(ii) Hence $∠ A B D = ∠ C A D$ . In $△ A B D$ , $∠ A B D + ∠ B A D = 9 0^{\circ}$ , so $∠ B A C = ∠ B A D + ∠ C A D = ∠ B A D + ∠ A B D = 9 0^{\circ}$ . (1)
(iii) $A D^{2} = 4 \times 9 = 36$ , so $A D = 6$ cm. $A B = 4^{2} + 6^{2} = 52 = 213$ cm and $A C = 9^{2} + 6^{2} = 117 = 313$ cm. Check: $A B^{2} + A C^{2} = 52 + 117 = 169 = 1 3^{2} = B C^{2}$ . (2)
Q5 · 5 marks · Long answerHarder board level
In $△ A B C$ , $∠ A = 9 0^{\circ}$ , $A B = 8$ cm, $A C = 6$ cm and $B C = 10$ cm. P is on AB with $A P = 3$ cm, and Q is on AC with $A Q = 4$ cm.
(i) Prove that $△ A P Q \sim △ A C B$ .
(ii) Find PQ.
(iii) Show that $∠ A P Q = ∠ A C B$ and $∠ A QP = ∠ A B C$ .
(iv) Find the perimeter of quadrilateral PBCQ.
Hint
Same-order ratios fail here, so pair AP with AC and AQ with AB.
Answer
(i) $\frac{A P}{A C} = \frac{3}{6} = \frac{1}{2}$ and $\frac{A Q}{A B} = \frac{4}{8} = \frac{1}{2}$ , with common $∠ A$ included between these sides, so $△ A P Q \sim △ A C B$ by SAS (2 marks). (ii) $\frac{P Q}{C B} = \frac{1}{2}$ , so $P Q = \frac{1}{2} \times 10 = 5$ cm. Check: $△ A P Q$ is right-angled at A and $3^{2} + 4^{2} = 5^{2}$ (1 mark). (iii) The correspondence is $A \leftrightarrow A$ , $P \leftrightarrow C$ , $Q \leftrightarrow B$ , so corresponding angles give $∠ A P Q = ∠ A C B$ and $∠ A QP = ∠ A B C$ (1 mark). (iv) $P B = 8 - 3 = 5$ cm, $B C = 10$ cm, $C Q = 6 - 4 = 2$ cm and $QP = 5$ cm, so the perimeter is $5 + 10 + 2 + 5 = 22$ cm (1 mark).
Q6 · 5 marks · Long answerHarder board level
In $△ A B C$ , D is on AB and E is on AC with $A D = 4$ cm, $D B = 8$ cm, $A E = 5$ cm, $E C = 10$ cm and $B C = 18$ cm.
(i) Prove that $D E ∥ B C$ .
(ii) Prove that $△ A D E \sim △ A B C$ and find DE.
(iii) BE and CD intersect at O. Prove that $△ O D E \sim △ O C B$ and find the ratio $O E : O B$ .
Hint
Use the converse of BPT for (i), SAS for (ii), and alternate angles from $D E ∥ B C$ with vertically opposite angles for (iii).
Answer
(i) $\frac{A D}{D B} = \frac{4}{8} = \frac{1}{2}$ and $\frac{A E}{E C} = \frac{5}{10} = \frac{1}{2}$ , so by the converse of the Basic Proportionality Theorem $D E ∥ B C$ (1½ marks). (ii) $A B = 12$ cm and $A C = 15$ cm, so $\frac{A D}{A B} = \frac{4}{12} = \frac{1}{3}$ and $\frac{A E}{A C} = \frac{5}{15} = \frac{1}{3}$ , with common $∠ A$ included, so $△ A D E \sim △ A B C$ by SAS, and $D E = \frac{1}{3} \times 18 = 6$ cm (1½ marks). (iii) Since $D E ∥ B C$ , $∠ O D E = ∠ O C B$ (alternate angles, transversal DC), and $∠ D O E = ∠ C O B$ (vertically opposite angles), so $△ O D E \sim △ O C B$ by AA similarity. Then $\frac{O E}{O B} = \frac{D E}{C B} = \frac{6}{18} = \frac{1}{3}$ , so $O E : O B = 1 : 3$ (2 marks).
Q7 · 4 marks · Case-basedHOTS / competency
Suppose your school makes a triangular rangoli ABC with $A B = 60$ cm, $A C = 45$ cm and $B C = 75$ cm. A smaller triangle ADE is to be filled with a different colour, with D on AB and E on AC. The designer wants $△ A D E \sim △ A C B$ , and decides that AE will be 5 cm longer than AD. Let $A D = x$ cm and $A E = (x + 5)$ cm.
(i) Which two ratios must be equal for $△ A D E \sim △ A C B$ by SAS? (1)
(ii) Find $x$ , AD and AE. (1)
(iii) Find DE, and decide with a reason whether the border DE is parallel to BC. (2)
Hint
D corresponds to C and E corresponds to B, so compare AD with AC and AE with AB. For (iii), test $\frac{A D}{D B}$ against $\frac{A E}{E C}$ .
Answer
(i) $\frac{A D}{A C} = \frac{A E}{A B}$ , with the common $∠ A$ included (1 mark). (ii) $\frac{x}{45} = \frac{x + 5}{60}$ gives $60 x = 45 x + 225$ , so $x = 15$ . Then $A D = 15$ cm and $A E = 20$ cm (1 mark). (iii) The ratio is $\frac{15}{45} = \frac{20}{60} = \frac{1}{3}$ , so $\frac{D E}{C B} = \frac{1}{3}$ and $D E = 25$ cm (1 mark). $D B = 45$ cm and $E C = 25$ cm, so $\frac{A D}{D B} = \frac{1}{3}$ but $\frac{A E}{E C} = \frac{4}{5}$ . If DE were parallel to BC, the Basic Proportionality Theorem would make these equal, so DE is not parallel to BC (1 mark).
Q8 · 3 marks · Short answerHOTS / competency
Kabir writes: "In $△ A B C$ , $A B = 4$ cm, $B C = 6$ cm and $∠ A = 5 0^{\circ}$ . In $△ P QR$ , $P Q = 6$ cm, $QR = 9$ cm and $∠ P = 5 0^{\circ}$ . Since $\frac{A B}{P Q} = \frac{B C}{QR}$ and $∠ A = ∠ P$ , the triangles are similar by SAS."
(i) Find the error, and state which angle would have to be equal for his conclusion to be valid.
(ii) Suppose that angle is equal and $A C = 5$ cm. Find PR and the ratio of the perimeters of the two triangles.
Hint
Which angle is formed by sides AB and BC? Once the triangles are similar, every pair of corresponding sides is in the same ratio.
Answer
(i) The ratios are both $\frac{2}{3}$ , but $∠ A$ is not included between AB and BC; the included angle is $∠ B$ . This is an SSA situation, which does not guarantee similarity. The conclusion would be valid if $∠ B = ∠ Q$ . (1½)
(ii) Then $△ A B C \sim △ P QR$ with ratio $\frac{2}{3}$ , so $P R = \frac{3}{2} \times 5 = 7.5$ cm. Perimeters: $4 + 6 + 5 = 15$ cm and $6 + 9 + 7.5 = 22.5$ cm, ratio $15 : 22.5 = 2 : 3$ , the same as the ratio of the sides. (1½)

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 SAS similarity criterion.Show answer
If one angle of a triangle equals one angle of another and the sides including these angles are proportional, the triangles are similar.
2.Is SAS similarity proved in your syllabus?Show answer
No. It is stated without proof. Only BPT is proved.
3.Is SSA a similarity criterion?Show answer
No. The equal angle must be between the proportional sides.
4.With $∠ A = ∠ P$ , which ratios must be equal?Show answer
$\frac{A B}{P Q}$ and $\frac{A C}{P R}$ .
5. $\frac{A D}{A C} = \frac{A E}{A B}$ with common angle A. Write the similarity.Show answer
$△ A D E \sim △ A C B$ .

SAS Similarity Criterion: frequently asked questions

How is SAS similarity different from SAS congruence?

SAS congruence needs the two sides to be equal. SAS similarity needs them only to be in the same ratio. In both, the equal angle must be between the sides. Congruence is simply the case where the ratio is 1.

Do I need to prove the SAS similarity criterion in the exam?

No. In the 2026-27 syllabus it is stated without proof, so you quote it and use it. The only theorem you must prove in this chapter is the Basic Proportionality Theorem.

How do I choose between AA, SSS and SAS?

Look at what is given. Two pairs of equal angles means AA, all six sides means SSS, and one equal angle with two sides of each triangle means SAS. Always check that the angle is included before using SAS.

Is the theorem on areas of similar triangles in the syllabus?

No. The ratio of areas of similar triangles is not in the 2026-27 syllabus, so do not use it in your answers.