MathsCBSE Class 9 · Triangles: Congruence Theorems

SAS, ASA and AAS Congruence and CPCT: Class 9

Updated 17 October 2026 · Fact-checked

Two triangles are congruent if their corresponding sides and angles are equal. Three rules prove it: SAS (two sides and the included angle), ASA (two angles and the included side) and AAS (two angles and any corresponding side). Once congruence is proved, CPCT (corresponding parts of congruent triangles) gives all remaining equalities. Name the triangles with vertices in corresponding order, and quote the rule.

Understand SAS, ASA and AAS Congruence with CPCT

Two triangles are congruent if one can be placed exactly on the other: all three pairs of corresponding sides and all three pairs of corresponding angles are equal. We write $△ A B C ≅ △ P QR$ , and the order of the letters matters: $A \leftrightarrow P$ , $B \leftrightarrow Q$ , $C \leftrightarrow R$ . Triangles are rigid: three given measurements often fix the whole triangle, which is why a few equalities are enough to prove congruence.

SAS rule (axiom). Two triangles are congruent if two sides and the included angle of one are equal to two sides and the included angle of the other. For $A B = P Q = 6$ , $∠ B = ∠ Q = 5 0^{\circ}$ and $B C = QR = 8$ , we get $△ A B C ≅ △ P QR$ . The angle must lie *between* the two sides.

ASA rule. Two triangles are congruent if two angles and the included side of one are equal to the corresponding parts of the other.

AAS rule. Two triangles are congruent if two angles and a non-included side are equal: since the angle sum is $18 0^{\circ}$ , the third angles are equal too, and the situation reduces to ASA. With $∠ A = ∠ D = 4 0^{\circ}$ , $∠ B = ∠ E = 6 5^{\circ}$ and $B C = E F$ , the third angles are both $7 5^{\circ}$ and the triangles are congruent by AAS. If instead the equal side is not in the matching position (say $A B = E F$ ), nothing can be concluded.

CPCT. Once two triangles are congruent, their corresponding parts (sides and angles) are equal. This is how congruence proves other facts. Always write: " $△ A B C ≅ △ P QR$ (SAS), so $A C = P R$ (CPCT)".

Typical uses.
1. *Diagonals meeting at midpoints:* if $A D$ and $B C$ bisect each other at $O$ , then $△ A O B ≅ △ D O C$ by SAS ( $A O = D O$ , $∠ A O B = ∠ D O C$ as vertically opposite angles, $B O = C O$ ); so $A B = D C$ and the alternate angles are equal, hence $A B ∥ D C$ .
2. *Diagonals of a parallelogram:* in parallelogram $A B C D$ with diagonals meeting at $O$ , $△ A O B ≅ △ C O D$ by ASA ( $∠ O A B = ∠ O C D$ and $∠ O B A = ∠ O D C$ as alternate angles, $A B = C D$ ), giving $A O = C O$ and $B O = D O$ .
3. *Perpendiculars to a diagonal:* in a parallelogram $A B C D$ , if $B E ⊥ A C$ and $D F ⊥ A C$ , then $△ A B E ≅ △ C D F$ by AAS ( $∠ A E B = ∠ C F D = 9 0^{\circ}$ , $∠ B A E = ∠ D C F$ as alternate angles, $A B = C D$ ) and so $B E = D F$ .
4. *Equal sides from a common angle:* in $△ A B C$ with $A B = A C$ and points $D$ on $A B$ , $E$ on $A C$ with $A D = A E$ , $△ A B E ≅ △ A C D$ by SAS (the angle at $A$ is common), so $B E = C D$ .

Measuring a river's width. Stand at $A$ opposite a tree $T$ across the river. Walk along the bank to $B$ , then the same distance again to $C$ . Turn at right angles and walk away from the river until $B$ lines up with the tree at $D$ . Then $△ A B T ≅ △ C B D$ by ASA ( $∠ A = ∠ C = 9 0^{\circ}$ , $A B = C B$ , $∠ A B T = ∠ C B D$ as vertically opposite angles), so $A T = C D$ : measuring $C D$ gives the river width.

Writing a proof. (1) Draw and mark equal parts. (2) List three matching pairs in order with a reason for each. (3) State the rule. (4) Apply CPCT to reach the conclusion.

SAS, ASA and AAS Congruence with CPCT: formulas to remember

Congruence notation: $△ A B C ≅ △ P QR \Rightarrow A \leftrightarrow P, B \leftrightarrow Q, C \leftrightarrow R$; Order of vertices matters.
SAS: $A B = P Q, ∠ B = ∠ Q, B C = QR$; The angle is included between the two sides.
ASA: $∠ B = ∠ Q, B C = QR, ∠ C = ∠ R$; The side is included between the two angles.
AAS: $∠ A = ∠ P, ∠ B = ∠ Q, B C = QR$; A non-included side; reduces to ASA.
CPCT: $corresponding parts of congruent triangles are equal$; Use after proving congruence.
Angle sum: $∠ A + ∠ B + ∠ C = 18 0^{\circ}$; Third angles are equal when two pairs are.

How to solve SAS, ASA and AAS Congruence with CPCT questions

Match three parts in order, name the rule, then use CPCT.

1Mark the given equal sides and angles on the figure; look for common sides and vertically opposite angles.
2Choose the triangles to compare and write the vertices in corresponding order.
3List three pairs of equal parts with a reason beside each (given, common, vertically opposite, alternate, bisector).
4Check the rule: SAS needs the included angle, ASA needs the included side, AAS needs any matching side.
5Write " $△ \dots ≅ △ \dots$ (rule)".
6Conclude with CPCT and finish the question (parallel lines, midpoints, lengths).

SAS, ASA and AAS Congruence with CPCT 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 congruence

Suppose a tailor cuts two triangular flags, each with sides 40 cm and 30 cm enclosing an angle of 60 degrees.

Will the two flags be identical in shape and size, and why?

Try it first, then show the solutionSolution

Two sides and the included angle of one flag equal those of the other: $40 = 40$ , $6 0^{\circ} = 6 0^{\circ}$ , $30 = 30$ .
By the SAS rule the flags are congruent, so the third sides and the other angles are also equal.

Answer: Yes, the flags are congruent by SAS, so their third sides and remaining angles match.

Memory hook: Two sides and the angle between them fix the triangle.

Tricks to remember SAS, ASA and AAS Congruence with CPCT

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

Included means between
SAS: the angle sits between the two sides. ASA: the side sits between the two angles.
Use it when: Choosing the rule.
Name in order
Write the second triangle's vertices in the same order as the matching ones in the first.
Use it when: Avoiding wrong CPCT.
Common side, vertical angles
Look first for a shared side or vertically opposite angles: they are free equalities.
Use it when: Starting a proof.

SAS, ASA and AAS Congruence with CPCT: solved examples

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

Example 1: SAS and its consequences · 3 marks · Short answerBoard level

In $△ A B C$ and $△ P QR$ , $A B = P Q = 6$ cm, $∠ B = ∠ Q = 5 0^{\circ}$ and $B C = QR = 8$ cm. Prove that the triangles are congruent. If $∠ A = 7 0^{\circ}$ , find $∠ R$ .

Try it first, then show the solutionSolution

$A B = P Q$ , $∠ B = ∠ Q$ (included angle), $B C = QR$ . (1)
So $△ A B C ≅ △ P QR$ (SAS). (1)
$∠ P = ∠ A = 7 0^{\circ}$ (CPCT), so $∠ R = ∠ C = 18 0^{\circ} - 5 0^{\circ} - 7 0^{\circ} = 6 0^{\circ}$ . (1)

Answer: The triangles are congruent by SAS and $∠ R = 6 0^{\circ}$ .

Example 2: Lines bisecting each other · 3 marks · Short answerBoard level

Segments $A D$ and $B C$ intersect at $O$ , which is the midpoint of both. Prove that $A B = D C$ and $A B ∥ D C$ .

Try it first, then show the solutionSolution

In $△ A O B$ and $△ D O C$ : $A O = D O$ (given), $∠ A O B = ∠ D O C$ (vertically opposite), $B O = C O$ (given). (1)
So $△ A O B ≅ △ D O C$ (SAS) and $A B = D C$ (CPCT). (1)
$∠ O A B = ∠ O D C$ (CPCT) and these are alternate angles for $A B$ and $D C$ cut by $A D$ , so $A B ∥ D C$ . (1)

Answer: $A B = D C$ and $A B ∥ D C$ .

Example 3: Perpendiculars to a diagonal · 5 marks · Long answerHarder board level

$A B C D$ is a parallelogram; $B E$ and $D F$ are perpendiculars from $B$ and $D$ to the diagonal $A C$ . Prove that $B E = D F$ . If $A C = 14$ cm and $B E = 6.4$ cm, find the area of the parallelogram.

Try it first, then show the solutionSolution

In $△ A B E$ and $△ C D F$ : $∠ A E B = ∠ C F D = 9 0^{\circ}$ . (1)
$∠ B A E = ∠ D C F$ (alternate angles, $A B ∥ D C$ with transversal $A C$ ). (1)
$A B = C D$ (opposite sides of a parallelogram). (½)
So $△ A B E ≅ △ C D F$ (AAS), and $B E = D F$ (CPCT). (1)
$△ A B C$ has base $A C = 14$ and height $B E = 6.4$ : area $= 44.8$ ; the diagonal splits the parallelogram into two equal triangles, so the area $= 2 \times 44.8 = 89.6$ cm $^{2}$ . (1½)

Answer: $B E = D F$ ; the parallelogram has area $89.6$ cm $^{2}$ .

Example 4: Which rule applies? · 3 marks · Short answerHarder board level

Two triangles $A B C$ and $D E F$ have $∠ A = ∠ D = 4 0^{\circ}$ and $∠ B = ∠ E = 6 5^{\circ}$ . State whether each extra condition proves congruence, and which rule it gives: (i) $A B = D E = 7$ cm; (ii) $B C = E F = 7$ cm; (iii) $A B = E F = 7$ cm.

Try it first, then show the solutionSolution

(i) $A B$ lies between $∠ A$ and $∠ B$ , and so does $D E$ between $∠ D$ and $∠ E$ : ASA, congruent. (1)
(ii) The third angles are $18 0^{\circ} - 4 0^{\circ} - 6 5^{\circ} = 7 5^{\circ}$ in both; $B C$ is opposite $∠ A$ and $E F$ is opposite $∠ D$ , so they correspond: AAS, congruent. (1)
(iii) $A B$ corresponds to $D E$ , not to $E F$ , so the equal sides are not corresponding sides and nothing can be concluded. (1)

Answer: (i) ASA; (ii) AAS; (iii) no conclusion.

Example 5: River width · 4 marks · Case-basedHOTS / competency

Case: to find the width of a river, a surveyor stands at $A$ opposite a tree $T$ , walks 40 m along the bank to $B$ and another 40 m to $C$ , then turns at right angles and walks away from the river until $B$ is in line with the tree, reaching $D$ . (i) Prove that $△ A B T ≅ △ C B D$ . (ii) Name the rule. (iii) If $C D = 57$ m, find the width of the river. (iv) Why do we need $A B = B C$ ?

Try it first, then show the solutionSolution

(i) $∠ T A B = ∠ D C B = 9 0^{\circ}$ (walking at right angles to the bank), $A B = C B = 40$ m, and $∠ A B T = ∠ C B D$ (vertically opposite angles, since $T$ , $B$ , $D$ are in a straight line). (1½)
(ii) These are two angles and the included side: ASA. (½)
(iii) $A T = C D$ by CPCT, so the river is $57$ m wide. (1)
(iv) The equal side $A B = C B$ is the included side in ASA; without it the triangles would have different sizes and $C D$ would not equal $A T$ . (1)

Answer: (i) proved; (ii) ASA; (iii) 57 m; (iv) it makes the triangles the same size.

Where students lose marks in SAS, ASA and AAS Congruence with CPCT

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

Mistake: Using a non-included angle in SAS
The equal angle is not between the equal sides.
Wrong working
$A B = P Q$ , $B C = QR$ , $∠ A = ∠ P$ gives congruence by SAS.
Right working
The angle must be $∠ B = ∠ Q$ , between the two sides.
Do this: Check that the angle is between the two sides.
Usually costs: The proof.
Mistake: Wrong order of vertices
Writing the triangles in non-corresponding order.
Wrong working
$△ A B C ≅ △ QP R$ for $A \leftrightarrow P$ .
Right working
$△ A B C ≅ △ P QR$ .
Do this: Match vertices as you list equal parts.
Usually costs: 1 mark.
Mistake: No reason for an equality
Writing equal parts without saying why.
Wrong working
$∠ A O B = ∠ D O C$ .
Right working
$∠ A O B = ∠ D O C$ (vertically opposite angles).
Do this: Give a reason for every equality.
Usually costs: ½ mark each.

Exam tips

List the three pairs in the order of the rule, with a reason beside each.
Look for common sides, vertically opposite angles and alternate angles first.
Never use AAA or SSA as a congruence rule.
End the proof with CPCT and then answer what was asked.

Practice questions on SAS, ASA and AAS Congruence with CPCT

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

Q1 · 1 mark · MCQBoard level
In $△ A B C$ and $△ D E F$ , $A B = D E$ , $B C = E F$ and $∠ B = ∠ E$ . The triangles are congruent by
(a) SAS
(b) ASA
(c) AAS
(d) SSS
Hint
The equal angle lies between the two equal sides.
Answer
(a) SAS. The angle $B$ is included between $A B$ and $B C$ , and angle $E$ between $D E$ and $E F$ .
Q2 · 3 marks · Short answerBoard level
In $△ P QR$ and $△ X Y Z$ , $∠ P = ∠ X = 8 0^{\circ}$ , $∠ Q = ∠ Y = 5 5^{\circ}$ and $QR = Y Z = 9$ cm. Prove congruence, name the rule and find $∠ R$ and the side equal to $P R$ .
Hint
The equal side is opposite the equal angles at $P$ and $X$ .
Answer
$QR$ is opposite $∠ P$ and $Y Z$ is opposite $∠ X$ : two angles and a corresponding non-included side, so $△ P QR ≅ △ X Y Z$ (AAS). $∠ R = 18 0^{\circ} - 8 0^{\circ} - 5 5^{\circ} = 4 5^{\circ} = ∠ Z$ . $P R = X Z$ by CPCT.
Q3 · 3 marks · Short answerHarder board level
$△ A B C$ and $△ D C B$ stand on the same base $B C$ on the same side, with $A B = D C$ and $∠ A B C = ∠ D C B$ . Prove that $A C = D B$ .
Hint
$B C$ is a common side.
Answer
In $△ A B C$ and $△ D C B$ : $A B = D C$ (given), $∠ A B C = ∠ D C B$ (given, included between the equal sides and $B C$ ) and $B C = C B$ (common). So $△ A B C ≅ △ D C B$ (SAS), and $A C = D B$ (CPCT).
Q4 · 5 marks · Long answerHarder board level
Prove that the diagonals of a parallelogram bisect each other. If the diagonals of parallelogram $A B C D$ meet at $O$ with $A C = 14$ cm and $B D = 18$ cm, find $A O$ and $B O$ .
Hint
Use triangles $A O B$ and $C O D$ and the alternate angles.
Answer
In $△ A O B$ and $△ C O D$ : $∠ O A B = ∠ O C D$ (alternate angles, $A B ∥ D C$ , transversal $A C$ ), $∠ O B A = ∠ O D C$ (alternate angles, transversal $B D$ ) and $A B = C D$ (opposite sides). So $△ A O B ≅ △ C O D$ (ASA), hence $A O = C O$ and $B O = D O$ (CPCT): the diagonals bisect each other. $A O = \frac{14}{2} = 7$ cm and $B O = \frac{18}{2} = 9$ cm.
Q5 · 4 marks · Case-basedHOTS / competency
Case: a signboard frame is an isosceles triangle $A B C$ with $A B = A C = 12$ cm. Points $D$ on $A B$ and $E$ on $A C$ are chosen with $A D = A E = 5$ cm, and the braces $B E$ and $C D$ are fixed. (i) Prove $B E = C D$ . (ii) Find $B D$ and $C E$ . (iii) If $∠ A B E = 2 2^{\circ}$ , find $∠ A C D$ . (iv) Prove that $△ D B C ≅ △ E C B$ .
Hint
Use SAS with the common angle $A$ ; then use base angles for (iv).
Answer
(i) In $△ A B E$ and $△ A C D$ : $A B = A C$ , $∠ A$ common, $A E = A D$ ; congruent by SAS, so $B E = C D$ (CPCT). (ii) $B D = 12 - 5 = 7$ cm and $C E = 12 - 5 = 7$ cm. (iii) $∠ A C D = ∠ A B E = 2 2^{\circ}$ (CPCT). (iv) In $△ D B C$ and $△ E C B$ : $D B = E C = 7$ , $∠ D B C = ∠ E C B$ (base angles of the isosceles triangle $A B C$ ), $B C$ common; so they are congruent (SAS).

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 angle does SAS use?Show answer
The angle included between the two sides.
2.What does CPCT stand for?Show answer
Corresponding parts of congruent triangles.
3.Is AAS a valid rule?Show answer
Yes, two angles and any corresponding side.
4.Name two equalities you get for free in a figure.Show answer
Common sides and vertically opposite angles.
5.Third angles when two pairs are equal?Show answer
Equal, by the angle sum property.

SAS, ASA and AAS Congruence with CPCT: frequently asked questions

Why is AAS valid when ASA is the rule?

Two equal angles force the third angles to be equal (angle sum 180^\circ), so a non-included side can be treated as an included one.

Do I always have to name the triangles in order?

Yes. CPCT uses the order of the letters to say which parts correspond.

What if the figure does not show equal marks?

Mark them yourself from the given information; a proof depends only on what is given or proved.