MathsCBSE Class 10 · Triangles

AAA and AA Similarity Criterion for Class 10 Triangles

Updated 14 October 2026 · Fact-checked

The AAA criterion says that if the corresponding angles of two triangles are equal, their corresponding sides are proportional and the triangles are similar. Since the angles add up to 180 degrees, two equal angles are enough, which is called AA. The board asks you to prove triangles similar by AA and then use side ratios.

Match the coloured arcs: two equal angles force the third to match, so the triangles are similar and every pair of sides has the same ratio.

Illustration drawn for QuestionBuddy.

Understand AAA and AA Similarity Criterion

Two triangles are similar when their corresponding angles are equal and their corresponding sides are in the same ratio. Checking all six facts is slow. The AAA criterion (NCERT Theorem 6.3, stated without proof) says that the angles alone are enough: if $∠ A = ∠ P$ , $∠ B = ∠ Q$ and $∠ C = ∠ R$ , then $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{C A}{R P}$ and $△ A B C \sim △ P QR$ .

Why should angles fix the sides? This is only a feel for why the result is true: the criterion is stated without proof in your syllabus, so do not write this in answers. Suppose $∠ A = ∠ P$ , $∠ B = ∠ Q$ , $∠ C = ∠ R$ and $△ P QR$ is the smaller triangle (if it is not, swap the roles of the two triangles).

Mark D on AB and E on AC so that $A D = P Q$ and $A E = P R$ .
Since $∠ A = ∠ P$ , $△ A D E ≅ △ P QR$ by SAS, so $∠ A D E = ∠ Q = ∠ B$ .
$∠ A D E$ and $∠ B$ are corresponding angles, so $D E ∥ B C$ .
BPT in $△ A B C$ gives $\frac{A D}{D B} = \frac{A E}{E C}$ , and so $\frac{A D}{A B} = \frac{A E}{A C}$ , that is, $\frac{P Q}{A B} = \frac{P R}{A C}$ .

Repeating the same idea at vertex B gives the third pair of sides in the same way, so all three ratios are equal. That is the link between this criterion and BPT.

You never need to check all three angles. The angle sum of a triangle is $18 0^{\circ}$ , so if two angles match, the third is $18 0^{\circ}$ minus the same two numbers and must match too. This shortened test is AA similarity, the criterion you will use most in the chapter.

Be careful: this works only for triangles. Two quadrilaterals can have all angles equal and still not be similar, for example a square and a non-square rectangle. For triangles, equal angles and proportional sides always come together.

In board problems the equal angles are rarely given directly. You find them from:

a common angle shared by both triangles
vertically opposite angles where two lines cross
alternate or corresponding angles from parallel lines
right angles from perpendiculars, altitudes or vertical objects
angles found from an angle sum or a given expression

Once AA is proved, write the similarity statement with matching vertices in the same order. Then read off the side ratio you need, cross-multiply, and you get results like $P A \cdot P B = P C \cdot P D$ or a missing length. Remember that AA only gives equal angles as input: the output is proportional sides, not equal sides.

AAA and AA Similarity Criterion: formulas to remember

AAA similarity criterion: $∠ A = ∠ P, ∠ B = ∠ Q, ∠ C = ∠ R \Rightarrow \frac{A B}{P Q} = \frac{B C}{QR} = \frac{C A}{R P}$; Stated without proof (NCERT Theorem 6.3). Conclusion: $△ A B C \sim △ P QR$ .
AA similarity criterion: $∠ A = ∠ P, ∠ B = ∠ Q \Rightarrow △ A B C \sim △ P QR$; Third angle follows from the angle sum $18 0^{\circ}$ .
Angle sum of a triangle: $∠ A + ∠ B + ∠ C = 18 0^{\circ}$; Reason why two equal angles are enough.
Line parallel to a side: $D E ∥ B C \Rightarrow △ A D E \sim △ A B C \Rightarrow \frac{D E}{B C} = \frac{A D}{A B}$; Corresponding angles equal plus common angle A. Use the whole side AB.
Intersecting chords-type product: $△ P A C \sim △ P D B \Rightarrow \frac{P A}{P D} = \frac{P C}{P B} \Rightarrow P A \cdot P B = P C \cdot P D$; Chords AB and CD meet at P inside a circle; $∠ P A C = ∠ P D B$ (angles in the same segment). Cross-multiply the ratio after proving AA.
Altitude on hypotenuse: $∠ B A C = 9 0^{\circ}, A D ⊥ B C \Rightarrow △ D B A \sim △ A B C \sim △ D A C, A D^{2} = B D \cdot D C$; Each smaller triangle shares one acute angle with the big one and has a right angle.

How to solve AAA and AA Similarity Criterion questions

Use this method for any question that needs AA: finding a length, proving a ratio or product, or a case study.

1Draw a rough figure and mark every given angle, parallel line and right angle.
2Pick the two triangles that contain all the lengths in the result you need.
3Find the first pair of equal angles and write its reason (common, vertically opposite, alternate, corresponding, $9 0^{\circ}$ ).
4Find the second pair of equal angles with its reason.
5Write 'Therefore, by AA similarity' and the statement with vertices in matching order.
6Write the ratio of corresponding sides, reading pairs of letters in the same positions.
7Cross-multiply or substitute and solve; give units and check the answer is sensible.

AAA and AA 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.

AA similarity criterion

Imagine you want to estimate the height of your school building without climbing it. You stand 40 m from the building on level ground and hold a ruler vertically 50 cm in front of your eye. Looking along the ruler, the line from your eye to the top of the building crosses the ruler 20 cm above your eye level. Your eye is 1.5 m above the ground.

Prove that the two triangles formed (eye, ruler, building) are similar, and find the height of the building. If each floor of the building is 3.5 m high, how many floors does it have?

Try it first, then show the solutionSolution

Let E be your eye, R the point on the ruler at eye level, T the point 20 cm above R, B the point on the building at eye level and P the top of the building.
$∠ E R T = ∠ E B P = 9 0^{\circ}$ (ruler and building are both vertical, line EB is horizontal).
$∠ T E R = ∠ P E B$ (common angle at the eye).
So $△ E R T \sim △ E B P$ by AA similarity.
$\frac{B P}{R T} = \frac{E B}{E R}$ , so $\frac{B P}{0.2} = \frac{40}{0.5} = 80$ .
$B P = 0.2 \times 80 = 16$ m (height above eye level).
Height of building $= 16 + 1.5 = 17.5$ m.
Number of floors $= \frac{17.5}{3.5} = 5$ .

Answer: The building is 17.5 m tall and has 5 floors.

Memory hook: The ruler trick: right angle at both and a common angle at your eye, so AA gives the height.

Tricks to remember AAA and AA Similarity Criterion

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

Two is enough
Triangle angles add to 180, so two matches force the third: AA = AAA.
Use it when: Never waste time proving the third angle.
CVAR checklist
Common, Vertically opposite, Alternate/corresponding, Right angle: the four places equal angles hide.
Use it when: Scan the figure in this order when looking for two equal angles.
Letters line up
First with first, second with second: the vertex order in the statement is the side map.
Use it when: Read ratios like AB/PQ straight from the statement.
Only triangles
Equal angles fix the shape only for triangles; a rectangle and a square share angles but not shape.
Use it when: Answers MCQs and assertion-reason on counter examples.

AAA and AA Similarity Criterion: solved examples

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

Example 1: Find a side using vertically opposite angles · 3 marks · Short answerBoard level

Line segments AD and BC intersect at O, with $∠ O A B = ∠ O D C$ . $A O = 6$ cm, $O D = 9$ cm, $A B = 8$ cm and $O B = 4$ cm. Prove that $△ A O B \sim △ D O C$ , and find CD and OC.

Try it first, then show the solutionSolution

In $△ A O B$ and $△ D O C$ : $∠ O A B = ∠ O D C$ (given). (½)
$∠ A O B = ∠ D O C$ (vertically opposite angles), so $△ A O B \sim △ D O C$ by AA. (½)
Hence $\frac{A O}{D O} = \frac{O B}{O C} = \frac{A B}{D C}$ , with scale factor $\frac{6}{9} = \frac{2}{3}$ . (½)
$\frac{2}{3} = \frac{8}{C D}$ , so $C D = 12$ cm. (½)
$\frac{2}{3} = \frac{4}{O C}$ , so $O C = 6$ cm. (½)
Check: the sides of $△ D O C$ (9, 6, 12) are $\frac{3}{2}$ times those of $△ A O B$ (6, 4, 8). (½)

Answer: $△ A O B \sim △ D O C$ by AA; CD = 12 cm and OC = 6 cm.

Example 2: MCQ: choose the correct similarity statement · 1 mark · MCQBoard level

In $△ A B C$ and $△ D E F$ , $∠ B = ∠ E = 6 5^{\circ}$ and $∠ C = ∠ D = 4 5^{\circ}$ . Which statement is correct?
(a) $△ A B C \sim △ D E F$
(b) $△ A B C \sim △ F E D$
(c) $△ A B C \sim △ E D F$
(d) $△ A B C \sim △ D F E$

Try it first, then show the solutionSolution

$∠ A = 18 0^{\circ} - 6 5^{\circ} - 4 5^{\circ} = 7 0^{\circ}$ and $∠ F = 18 0^{\circ} - 6 5^{\circ} - 4 5^{\circ} = 7 0^{\circ}$ (angle sum).
Matching equal angles: A with F, B with E, C with D.
So, by AA, $△ A B C \sim △ F E D$ . (1)

Answer: (b) $△ A B C \sim △ F E D$

Example 3: MCQ: ratio of sides with a parallel line · 1 mark · MCQBoard level

In $△ A B C$ , D lies on AB and E lies on AC with $D E ∥ B C$ . If $A D = 2$ cm, $D B = 3$ cm and $B C = 10$ cm, then DE equals
(a) 4 cm
(b) $\frac{20}{3}$ cm
(c) 5 cm
(d) 6 cm

Try it first, then show the solutionSolution

$∠ A$ 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.
$\frac{D E}{B C} = \frac{A D}{A B} = \frac{2}{2 + 3} = \frac{2}{5}$ (use the whole side AB).
$D E = \frac{2}{5} \times 10 = 4$ cm. (1)

Answer: (a) 4 cm

Example 4: Assertion-reason: equal angles in triangles and quadrilaterals · 1 mark · Assertion-reasonBoard level

Assertion (A): A square of side 4 cm and a rectangle of 4 cm by 6 cm are not similar, even though all their angles are equal.
Reason (R): If the corresponding angles of two triangles are equal, their corresponding sides are proportional.
(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

A: all angles are $9 0^{\circ}$ , but side ratios $\frac{4}{4} = 1$ and $\frac{4}{6} = \frac{2}{3}$ differ, so the figures are not similar. A is true.
R is the AAA criterion, which is true.
R is about triangles only; it does not explain why the quadrilaterals fail. The reason for A is that sides are not proportional. So the answer is (b). (1)

Answer: (b) Both A and R are true but R is not the correct explanation of A.

Example 5: Two-mark question: prove similar and find a side · 2 marks · Very short answerBoard level

P and Q are points on sides AB and AC of $△ A B C$ such that $∠ A P Q = ∠ A B C$ . If $A P = 4$ cm, $A B = 10$ cm and $B C = 7.5$ cm, find PQ.

Try it first, then show the solutionSolution

In $△ A P Q$ and $△ A B C$ : $∠ A$ is common. (½)
$∠ A P Q = ∠ A B C$ (given), so $△ A P Q \sim △ A B C$ by AA. (½)
$\frac{P Q}{B C} = \frac{A P}{A B}$ , so $P Q = \frac{4 \times 7.5}{10} = 3$ cm. (1)

Answer: PQ = 3 cm

Example 6: Product result from two chords · 3 marks · Short answerBoard level

Two chords AB and CD of a circle intersect at P inside the circle. Angles in the same segment are equal, so $∠ C A B = ∠ C D B$ , that is, $∠ P A C = ∠ P D B$ . Prove that $P A \cdot P B = P C \cdot P D$ . If $P A = 4$ cm, $P B = 9$ cm and $P C = 3$ cm, find CD.

Try it first, then show the solutionSolution

In $△ P A C$ and $△ P D B$ : $∠ P A C = ∠ P D B$ (angles in the same segment, given). (½)
$∠ A P C = ∠ D P B$ (vertically opposite angles). (½)
So $△ P A C \sim △ P D B$ by AA. (½)
$\frac{P A}{P D} = \frac{P C}{P B}$ , so $P A \cdot P B = P C \cdot P D$ . (½)
$4 \times 9 = 3 \times P D$ , so $P D = 12$ cm. (½)
$C D = P C + P D = 3 + 12 = 15$ cm. (½)

Answer: $P A \cdot P B = P C \cdot P D$ ; CD = 15 cm.

Example 7: Algebraic sides in a trapezium · 3 marks · Short answerHarder board level

ABCD is a trapezium with $A B ∥ D C$ . Its diagonals AC and BD meet at O. If $A O = (3 x - 1)$ cm, $O C = (5 x - 3)$ cm, $B O = (2 x + 1)$ cm and $O D = (6 x - 5)$ cm, prove that $\frac{A O}{O C} = \frac{B O}{O D}$ , find x and the lengths of both diagonals.

Try it first, then show the solutionSolution

In $△ A O B$ and $△ C O D$ : $∠ O A B = ∠ O C D$ (alternate angles, $A B ∥ D C$ ). (½)
$∠ A O B = ∠ C O D$ (vertically opposite), so $△ A O B \sim △ C O D$ by AA, and $\frac{A O}{O C} = \frac{B O}{O D}$ . (½)
$(3 x - 1) (6 x - 5) = (2 x + 1) (5 x - 3)$ , so $18 x^{2} - 21 x + 5 = 10 x^{2} - x - 3$ . (½)
$8 x^{2} - 20 x + 8 = 0$ , so $2 x^{2} - 5 x + 2 = 0$ . (½)
$(2 x - 1) (x - 2) = 0$ , so $x = \frac{1}{2}$ or $x = 2$ . (½)
$x = \frac{1}{2}$ gives $O C = - \frac{1}{2}$ , rejected. So $x = 2$ : AO = 5, OC = 7, BO = 5, OD = 7, giving AC = 12 cm and BD = 12 cm. (½)

Answer: x = 2; AC = 12 cm and BD = 12 cm (check: $\frac{5}{7} = \frac{5}{7}$ ).

Example 8: Altitude on the hypotenuse · 5 marks · Long answerHarder board level

In $△ A B C$ , $∠ B A C = 9 0^{\circ}$ and AD is perpendicular to BC, with D on BC. Prove that $△ D B A \sim △ D A C$ and hence $A D^{2} = B D \cdot D C$ . If $B D = 4$ cm and $D C = 9$ cm, find AD, AB and AC. (Pythagoras may be used as a known result.)

Try it first, then show the solutionSolution

Let $∠ B = x$ . In $△ A B D$ , $∠ B A D = 9 0^{\circ} - x$ (angle sum, $∠ A D B = 9 0^{\circ}$ ). (½)
In $△ A B C$ , $∠ C = 9 0^{\circ} - x$ (angle sum). (½)
In $△ D A C$ , $∠ D A C = 9 0^{\circ} - ∠ C = x$ . (½)
In $△ D B A$ and $△ D A C$ : $∠ B D A = ∠ A D C = 9 0^{\circ}$ and $∠ D B A = ∠ D A C = x$ . (½)
So $△ D B A \sim △ D A C$ by AA. (½)
$\frac{D B}{D A} = \frac{D A}{D C}$ , so $A D^{2} = B D \cdot D C$ . (1)
$A D^{2} = 4 \times 9 = 36$ , so $A D = 6$ cm. (½)
$A B^{2} = B D^{2} + A D^{2} = 16 + 36 = 52$ , so $A B = 213$ cm. (½)
$A C^{2} = A D^{2} + D C^{2} = 36 + 81 = 117$ , so $A C = 313$ cm. (½)

Answer: AD = 6 cm, AB = $213$ cm, AC = $313$ cm (check: $52 + 117 = 169 = 1 3^{2}$ , and BC = 13 cm).

Example 9: Perpendiculars in an isosceles triangle · 5 marks · Long answerHarder board level

In $△ A B C$ , $A B = A C$ (isosceles). E is a point on side CB produced beyond B. AD is perpendicular to BC with D on BC, and EF is perpendicular to AC with F on AC. Prove that $△ A B D \sim △ E C F$ and that $A B \cdot E F = A D \cdot E C$ . If $A B = 10$ cm, $B D = 6$ cm, $A D = 8$ cm and $E C = 15$ cm, find EF, CF, AF and EB.

Try it first, then show the solutionSolution

$A B = A C$ , so $∠ A B C = ∠ A C B$ (angles opposite equal sides). (½)
Thus $∠ A B D = ∠ E C F$ , since D lies on BC and E lies on CB produced, so $∠ E C F$ is $∠ A C B$ . (½)
$∠ A D B = ∠ E F C = 9 0^{\circ}$ (given perpendiculars). (½)
So $△ A B D \sim △ E C F$ by AA. (½)
$\frac{A B}{E C} = \frac{B D}{C F} = \frac{A D}{E F}$ . (½)
From $\frac{A B}{E C} = \frac{A D}{E F}$ , cross-multiplying gives $A B \cdot E F = A D \cdot E C$ . (½)
$10 \times E F = 8 \times 15$ , so $E F = 12$ cm. (½)
$\frac{10}{15} = \frac{6}{C F}$ , so $C F = 9$ cm. (½)
$A C = A B = 10$ cm, so $A F = A C - C F = 10 - 9 = 1$ cm. (½)
$△ A B D ≅ △ A C D$ (RHS), so $D C = B D = 6$ cm and $B C = 12$ cm; hence $E B = E C - B C = 15 - 12 = 3$ cm. (½)

Answer: EF = 12 cm, CF = 9 cm, AF = 1 cm and EB = 3 cm (check: $9^{2} + 1 2^{2} = 1 5^{2}$ ).

Example 10: Line through the diagonal point of a trapezium · 5 marks · Long answerHarder board level

ABCD is a trapezium with $A B ∥ D C$ , $A B = 12$ cm and $D C = 6$ cm. The diagonals AC and BD meet at O. A line through O parallel to AB meets AD at E and BC at F. Prove that $E O = O F$ , and find EF and the ratio $A E : E D$ .

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 $△ A O B \sim △ C O D$ by AA. (1)
$\frac{A O}{O C} = \frac{A B}{C D} = \frac{12}{6} = 2$ , so $\frac{A O}{A C} = \frac{2}{3}$ and $\frac{C O}{C A} = \frac{1}{3}$ . (½)
$E F ∥ A B$ and $A B ∥ D C$ , so $E O ∥ D C$ . (½)
In $△ A E O$ and $△ A D C$ : $∠ A$ common, $∠ A E O = ∠ A D C$ (corresponding angles, $E O ∥ D C$ ), so $△ A E O \sim △ A D C$ by AA. (½)
$\frac{E O}{D C} = \frac{A E}{A D} = \frac{A O}{A C} = \frac{2}{3}$ , so $E O = \frac{2}{3} \times 6 = 4$ cm and $A E : E D = 2 : 1$ . (1)
In $△ C O F$ and $△ C A B$ : $∠ C$ common, $∠ C O F = ∠ C A B$ (corresponding angles, $O F ∥ A B$ ), so $△ C O F \sim △ C A B$ by AA. (½)
$\frac{O F}{A B} = \frac{C O}{C A} = \frac{1}{3}$ , so $O F = \frac{1}{3} \times 12 = 4$ cm. (½)
Hence $E O = O F = 4$ cm and $E F = 8$ cm. (½)

Answer: EO = OF = 4 cm, EF = 8 cm and AE : ED = 2 : 1.

Example 11: Case study: shadows on sports day · 4 marks · Case-basedHOTS / competency

Imagine your school sports day. At 4 p.m., a vertical flagpole on level ground casts a shadow 18 m long, while a student 1.5 m tall standing nearby casts a shadow 2.7 m long.
(i) Why are the pole's triangle and the student's triangle similar? (1 mark)
(ii) Find the height of the flagpole. (1 mark)
(iii) In the evening, the same student stands some distance from a lamp post and casts a shadow 2 m long. She walks 3 m straight away from the post, and her shadow becomes 3.5 m long. Find how far she first stood from the post and the height of the lamp. (2 marks)

Try it first, then show the solutionSolution

(i) Both stand vertically, so each makes $9 0^{\circ}$ with the ground; the sun's rays make the same angle with the ground at that moment. So the triangles are similar by AA. (1)
(ii) $\frac{h}{1.5} = \frac{18}{2.7}$ , so $h = \frac{1.5 \times 18}{2.7} = 10$ m. (1)
(iii) Let the lamp be $H$ m high and her first distance from the post $d$ m. The lamp's top, the tip of her shadow and the foot of the post form a right triangle similar (AA: common angle at the shadow tip, right angles) to the one formed by the student and her shadow: $\frac{1.5}{H} = \frac{2}{d + 2}$ . (½)
Second position: her distance is $d + 3$ and the shadow tip is $d + 3 + 3.5$ from the post, so $\frac{1.5}{H} = \frac{3.5}{d + 6.5}$ . (½)
Equate: $\frac{2}{d + 2} = \frac{3.5}{d + 6.5}$ , so $2 d + 13 = 3.5 d + 7$ and $d = 4$ m. Then $H = \frac{1.5 \times 6}{2} = 4.5$ m. Check: $\frac{1.5}{4.5} = \frac{3.5}{10.5}$ . (1)

Answer: (i) AA: right angles and equal sun angle. (ii) 10 m. (iii) She first stood 4 m from the post, and the lamp is 4.5 m high.

Where students lose marks in AAA and AA Similarity Criterion

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

Mistake: Writing vertices in the wrong order
The statement order decides which sides pair up; a jumbled order gives wrong ratios.
Wrong working
$∠ P S T = ∠ P R Q$ , so $△ P S T \sim △ P QR$ and $\frac{P S}{P Q} = \frac{P T}{P R}$ .
Right working
S matches R, so $△ P S T \sim △ P R Q$ and $\frac{P S}{P R} = \frac{P T}{P Q}$ .
Do this: Match the equal angles first, then write both triangles in that order.
Usually costs: Usually the final answer and ratio marks.
Mistake: Giving no reason for equal angles
Board marking awards marks for each reason, not just the statement.
Wrong working
$∠ A O B = ∠ C O D$ , $∠ O A B = ∠ O C D$ , so similar.
Right working
$∠ A O B = ∠ C O D$ (vertically opposite); $∠ O A B = ∠ O C D$ (alternate angles, $A B ∥ D C$ ); so similar by AA.
Do this: Put the reason in brackets after every angle equation, and name the criterion.
Usually costs: ½ to 1 mark per missing reason.
Mistake: Using the part instead of the whole side
In $△ A D E \sim △ A B C$ , AD matches AB, not DB.
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: Sides of the similar triangles are whole sides of each triangle.
Usually costs: The answer mark and often the next part.
Mistake: Concluding equal sides from AA
AA gives proportional sides; equal sides need congruence.
Wrong working
$△ A O B \sim △ D O C$ , so $A B = D C$ .
Right working
$△ A O B \sim △ D O C$ , so $\frac{A B}{D C} = \frac{A O}{D O}$ .
Do this: After similarity, always write a ratio, never an equality of sides.
Usually costs: Loses the conclusion marks.

Exam tips

Write 'by AA similarity criterion' explicitly; the criterion name carries a mark in proofs.
In MCQs on correspondence, write the full similarity statement on rough paper before choosing an option.
When a result is a product like $P A \cdot P B = P C \cdot P D$ , choose the triangles whose sides are exactly those four lengths.
In case studies, the 1-mark 'why similar' part needs both angle reasons: the right angle and the equal or common angle.
If Pythagoras is needed after similarity, use it as a known result; do not try to prove it.

Practice questions on AAA and AA 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 · 3 marks · Short answerHarder board level
In $△ A B C$ , D is a point on BC with $B D = 6$ cm and $D C = 4$ cm. A line through D parallel to AB meets AC at E, and a line through D parallel to AC meets AB at F. If $A B = 15$ cm and $A C = 12$ cm, prove the two similarities you use, find DE and DF, and hence find the perimeter of the parallelogram AFDE.
Hint
$△ C D E \sim △ C B A$ and $△ B D F \sim △ B C A$ , each by AA with corresponding angles.
Answer
$D E ∥ A B$ : $∠ C$ common and $∠ C D E = ∠ C B A$ (corresponding), so $△ C D E \sim △ C B A$ and $\frac{D E}{A B} = \frac{C D}{C B} = \frac{4}{10}$ , giving $D E = 6$ cm. $D F ∥ A C$ : $∠ B$ common and $∠ B D F = ∠ B C A$ , so $△ B D F \sim △ B C A$ and $\frac{D F}{A C} = \frac{B D}{B C} = \frac{6}{10}$ , giving $D F = 7.2$ cm. AFDE has both pairs of opposite sides parallel, so its perimeter $= 2 (6 + 7.2) = 26.4$ cm.
Q2 · 3 marks · Short answerHarder board level
In $△ A B C$ , D is on AB and E is on AC such that $∠ A D E = ∠ A C B$ . If $A D = x$ cm, $D B = 5$ cm, $A E = 3$ cm, $E C = 9$ cm and $B C = 15$ cm, prove that $△ A D E \sim △ A C B$ , find x and find DE.
Hint
Angle A is common; D matches C and E matches B. The ratio gives a quadratic in x.
Answer
$∠ A$ common, $∠ A D E = ∠ A C B$ , so $△ A D E \sim △ A C B$ by AA. $\frac{A D}{A C} = \frac{A E}{A B}$ : $x (x + 5) = 3 \times 12 = 36$ , so $x^{2} + 5 x - 36 = 0$ , $(x + 9) (x - 4) = 0$ , x = 4 (x = −9 rejected). Ratio $= \frac{4}{12} = \frac{1}{3}$ , so $D E = \frac{1}{3} \times 15 = 5$ cm.
Q3 · 5 marks · Long answerHarder board level
In $△ A B C$ , altitudes AD and CE meet at H, with D on BC and E on AB. Prove that $△ A E H \sim △ C D H$ and $△ A B D \sim △ C B E$ . If $A B = 15$ cm, $B D = 9$ cm and $B C = 20$ cm, find BE.
Hint
Right angles at E and D, with vertically opposite angles at H; for the second pair, angle B is common.
Answer
$∠ A E H = ∠ C D H = 9 0^{\circ}$ , $∠ A H E = ∠ C H D$ (vertically opposite): AA. $∠ A D B = ∠ C E B = 9 0^{\circ}$ , $∠ B$ common: AA. $\frac{A B}{C B} = \frac{B D}{B E}$ : $\frac{15}{20} = \frac{9}{B E}$ , BE = 12 cm.
Q4 · 3 marks · Short answerHarder board level
In $△ A B C$ , D is a point on BC such that $∠ B A D = ∠ B C A$ . $A B = 6$ cm, $B D = x$ cm and $B C = (x + 5)$ cm. Show that $A B^{2} = B D \cdot B C$ , and find x, BC and DC.
Hint
Prove $△ B A D \sim △ B C A$ first; the equation in x is quadratic.
Answer
$∠ B$ common and $∠ B A D = ∠ B C A$ , so $△ B A D \sim △ B C A$ by AA and $\frac{B A}{B C} = \frac{B D}{B A}$ , giving $A B^{2} = B D \cdot B C$ . So $36 = x (x + 5)$ , $x^{2} + 5 x - 36 = 0$ , $(x + 9) (x - 4) = 0$ , and $x = 4$ (a length cannot be $- 9$ ). BC = 9 cm and DC = 5 cm.
Q5 · 5 marks · Long answerHOTS / competency
Two vertical poles AB = 6 m and CD = 3 m stand on level ground with feet B and D, 12 m apart. A rope joins the top A of the taller pole to the foot D of the shorter pole, and another rope joins the top C of the shorter pole to the foot B of the taller pole. The ropes cross at P, and PQ is perpendicular to the ground with Q on BD. Prove that $\frac{1}{P Q} = \frac{1}{A B} + \frac{1}{C D}$ , find PQ and BQ, and state whether PQ depends on the distance BD.
Hint
Use two separate AA pairs: $△ D QP$ with $△ D B A$ , and $△ B QP$ with $△ B D C$ . Then add the two ratios.
Answer
$∠ D$ common, $∠ D QP = ∠ D B A = 9 0^{\circ}$ , so $△ D QP \sim △ D B A$ and $\frac{P Q}{A B} = \frac{D Q}{D B}$ . $∠ B$ common, $∠ B QP = ∠ B D C = 9 0^{\circ}$ , so $△ B QP \sim △ B D C$ and $\frac{P Q}{C D} = \frac{B Q}{B D}$ . Adding: $P Q (\frac{1}{A B} + \frac{1}{C D}) = \frac{D Q + B Q}{B D} = 1$ , so $\frac{1}{P Q} = \frac{1}{A B} + \frac{1}{C D}$ . $\frac{1}{P Q} = \frac{1}{6} + \frac{1}{3} = \frac{1}{2}$ , so PQ = 2 m. $B Q = \frac{12 \times 2}{3} = 8$ m (and DQ = 4 m, total 12 m). PQ does not depend on BD.
Q6 · 3 marks · Short answerHarder board level
In $△ A B C$ , $∠ A C B = 9 0^{\circ}$ and CD is perpendicular to AB with D on AB. Prove that $B C^{2} = B D \cdot B A$ . If $A D = 5$ cm and $D B = 4$ cm, find BC and AC, and hence show that $\frac{B C ^{2}}{A C ^{2}} = \frac{B D}{A D}$ .
Hint
Compare $△ B D C$ with $△ B C A$ (angle B common), and $△ A D C$ with $△ A C B$ (angle A common).
Answer
$∠ B D C = ∠ B C A = 9 0^{\circ}$ , $∠ B$ common, so $△ B D C \sim △ B C A$ and $\frac{B D}{B C} = \frac{B C}{B A}$ , giving $B C^{2} = B D \cdot B A = 4 \times 9 = 36$ , BC = 6 cm. Similarly $△ A D C \sim △ A C B$ gives $A C^{2} = A D \cdot A B = 5 \times 9 = 45$ , AC = $35$ cm (check: $36 + 45 = 81 = 9^{2}$ ). Then $\frac{B C ^{2}}{A C ^{2}} = \frac{36}{45} = \frac{4}{5} = \frac{B D}{A D}$ ; in general $\frac{B C ^{2}}{A C ^{2}} = \frac{B D \cdot B A}{A D \cdot A B} = \frac{B D}{A D}$ .
Q7 · 1 mark · Assertion-reasonHOTS / competency
Assertion (A): Two rhombuses whose corresponding angles are equal are always similar.
Reason (R): Two polygons with the same number of sides are similar if their corresponding angles are equal and their corresponding sides are 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
Unlike a square and a rectangle, a rhombus has all four sides equal. What does that do to the side ratios?
Answer
(a). In each rhombus all four sides are equal, so the ratio of corresponding sides is the same for all four pairs (side of one ÷ side of the other). With equal angles too, both conditions in R hold, so A is true and R explains it. Compare a square and a rectangle: equal angles, but the side ratios differ, so they are not similar.

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 AAA similarity criterion.Show answer
If corresponding angles of two triangles are equal, their corresponding sides are proportional and the triangles are similar.
2.Why is AA enough?Show answer
The angle sum is $18 0^{\circ}$ , so the third angles are equal automatically.
3.Is the AAA criterion proved in your syllabus?Show answer
No, it is stated without proof.
4.Name four sources of equal angles.Show answer
Common angle, vertically opposite angles, alternate or corresponding angles, right angles.
5.Does AAA work for quadrilaterals?Show answer
No. A square and a non-square rectangle have equal angles but are not similar.
6.If $D E ∥ B C$ in $△ A B C$ , which triangles are similar?Show answer
$△ A D E \sim △ A B C$ by AA.

AAA and AA Similarity Criterion: frequently asked questions

What is the difference between AAA and AA similarity?

AAA needs all three pairs of angles equal, AA needs only two. They are really the same criterion, because the third pair of angles is forced to be equal by the angle sum property. In answers you can write AA similarity criterion.

Do I have to prove the AA criterion in the board exam?

No. The 2026-27 syllabus lists it as stated without proof. The only theorem you must prove in this chapter is the Basic Proportionality Theorem.

Can I use AA to prove triangles congruent?

No. AA only shows the triangles have the same shape, so their sides are in proportion. For congruence you need a side condition as well, as in ASA or AAS from Class 9.

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

No, the ratio of areas theorem is not in the 2026-27 syllabus. Use similar triangles only for ratios of corresponding sides (and so perimeters).

How do I know which sides correspond after proving AA?

Write the similarity statement so that equal angles sit in the same positions. Then the side formed by the first two letters of one triangle matches the side formed by the first two letters of the other, and so on.