What is CPCT in congruent triangles (Class 9)?

CPCT stands for Corresponding Parts of Congruent Triangles. Once two triangles are proved congruent using any criterion (SAS, AAS, RHS, SSS, or ASA), all their corresponding sides and angles become equal. CPCT is used in Exercise 7.1 to extract information such as BC = BD or BD = AC after the triangles are proved congruent.

How do you prove ΔABC ≅ ΔABD in Question 1 of Exercise 7.1?

In quadrilateral ACBD where AC = AD and AB bisects ∠A, consider ΔABC and ΔABD. The side AB is common, AC = AD (given), and ∠CAB = ∠DAB (AB bisects ∠A). These two sides and the included angle match, so by the SAS congruence rule, ΔABC ≅ ΔABD. By CPCT, BC = BD.

Why is RHS used in Question 9 of Exercise 7.1?

In Question 9, ΔBED and ΔCFD are right-angled triangles (∠DEB = ∠DFC = 90°). The hypotenuses BD = CD (D is the midpoint of BC) and one leg DE = DF (given). Since both triangles share a right angle, have equal hypotenuses and equal sides, the RHS (Right angle-Hypotenuse-Side) congruence rule applies: ΔBED ≅ ΔCFD.

How is CM = ½ AB proved in Question 6 of Exercise 7.1?

In Question 6, M is the midpoint of AB in right triangle ABC (right angle at C). Point D is chosen such that DM = CM. First, ΔAMC ≅ ΔBMD by SAS (AM = BM, CM = DM, vertically opposite angles). Then ΔDBC ≅ ΔACB by SAS, giving DC = AB by CPCT. Since DC = CM + MD = CM + CM = 2CM, we get 2CM = AB, hence CM = ½ AB.

Exercise 7.1 — Congruency Problems | Class 9 Maths | Triangles

What Is Congruence of Triangles?

Two triangles are said to be congruent when they have exactly the same shape and the same size — every side and every angle of one triangle matches perfectly with the corresponding side and angle of the other. When triangles ΔABC ≅ ΔPQR, it means AB = PQ, BC = QR, CA = RP, ∠A = ∠P, ∠B = ∠Q, and ∠C = ∠R.

In Exercise 7.1, students use three key congruence criteria — SAS, AAS, and RHS — together with the powerful principle of CPCT (Corresponding Parts of Congruent Triangles) to prove relationships between sides and angles in geometric figures involving quadrilaterals, isosceles triangles, and right triangles.

CPCT Principle: Once two triangles are proved congruent by any criterion, all their corresponding parts (sides and angles) are automatically equal. CPCT is the tool used to extract further conclusions from a proved congruence.

Congruence Criteria Used in Exercise 7.1

Rule	Full Name	Condition
SAS	Side – Angle – Side	Two sides and the included angle of one triangle equal the corresponding two sides and included angle of another.
AAS	Angle – Angle – Side	Two angles and one non-included side of one triangle equal the corresponding two angles and side of another.
RHS	Right angle – Hypotenuse – Side	In two right-angled triangles, the hypotenuse and one side of one triangle equal the hypotenuse and the corresponding side of the other.

Important Note: SSA (two sides and a non-included angle) is not a valid congruence criterion in general. RHS is the special case of SSA that works — only for right-angled triangles where the matching angle is 90°.

Visual Overview — SAS Congruence

SAS Congruence: Two equal sides (double-tick and green-tick) with the included angle ∠B = ∠P

Exercise 7.1 — All Problems with Solutions

Question 1

In quadrilateral ACBD, AC = AD and AB bisects ∠A. Show that ΔABC ≅ ΔABD. What can you say about BC and BD?

In ΔABC and ΔABD:

AC = AD (Given)

∠CAB = ∠DAB (AB bisects ∠A — Given)

AB = AB (Common side)

∴ ΔABC ≅ ΔABD (By SAS congruence rule)

∴ BC = BD (By CPCT) — B lies on the perpendicular bisector of CD.

Question 2

ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that (i) ΔABD ≅ ΔBAC (ii) BD = AC (iii) ∠ABD = ∠BAC

In ΔABD and ΔBAC:

AD = BC (Given)

∠DAB = ∠CBA (Given)

AB = AB (Common side)

(i) ∴ ΔABD ≅ ΔBAC (By SAS congruence rule)

(ii) BD = AC (By CPCT)

(iii) ∠ABD = ∠BAC (By CPCT)

Question 3

AD and BC are equal and perpendicular to a line segment AB. Show that CD bisects AB.

Let CD intersect AB at O. In ΔOAD and ΔOBC:

AD = BC (Given)

∠AOD = ∠BOC (Vertically opposite angles)

∠OAD = ∠OBC = 90° (AD ⊥ AB, BC ⊥ AB — Given)

∴ ΔOAD ≅ ΔOBC (By AAS congruence rule)

∴ OA = OB (By CPCT) → CD bisects AB at O

Question 4

l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that ΔABC ≅ ΔCDA.

Since l ∥ m and p ∥ q, ABCD is a parallelogram. In ΔABC and ΔCDA:

AB = CD (Opposite sides of parallelogram)

BC = AD (Opposite sides of parallelogram)

∠ABC = ∠CDA (Opposite angles of parallelogram)

∴ ΔABC ≅ ΔCDA (By SAS congruence rule)

Question 5

In the figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

In ΔABC and ΔADE:

AB = AD (Given)

AC = AE (Given)

∠BAC = ∠BAD + ∠DAC = ∠EAC + ∠DAC = ∠DAE (Since ∠BAD = ∠EAC)

∴ ΔABC ≅ ΔADE (By SAS congruence rule)

∴ BC = DE (By CPCT)

Question 6

In right triangle ABC (right angle at C), M is the midpoint of hypotenuse AB. C is joined to M and produced to D such that DM = CM. Point D is joined to B. Show that: (i) ΔAMC ≅ ΔBMD (ii) ∠DBC is a right angle (iii) ΔDBC ≅ ΔACB (iv) CM = ½ AB

(i) In ΔAMC and ΔBMD:

AM = BM (M is midpoint of AB)

∠AMC = ∠BMD (Vertically opposite angles)

CM = DM (Given)

∴ ΔAMC ≅ ΔBMD (By SAS)

(ii) ΔAMC ≅ ΔBMD → ∠ACM = ∠BDM (CPCT) → AC ∥ DB (alternate interior angles). Since BC is transversal: ∠DBC + ∠ACB = 180° → ∠DBC + 90° = 180°

∴ ∠DBC = 90° (∠DBC is a right angle)

(iii) In ΔDBC and ΔACB:

DB = AC (CPCT from (i))

∠DBC = ∠ACB = 90°

BC = BC (Common)

∴ ΔDBC ≅ ΔACB (By SAS)

(iv) ΔDBC ≅ ΔACB → DC = AB (CPCT)

CM + DM = AB → CM + CM = AB (DM = CM)

∴ 2CM = AB → CM = ½ AB

Question 7

ABCD is a square and ΔAPB is an equilateral triangle. Prove that ΔAPD ≅ ΔBPC.

In ΔAPD and ΔBPC:

AP = BP (Sides of equilateral ΔAPB)

AD = BC (Sides of square ABCD)

∠PAD = ∠PBC = 90° − 60° = 30° (∠DAB = 90°, ∠PAB = 60°)

∴ ΔAPD ≅ ΔBPC (By SAS congruence rule)

Question 8

ΔABC is isosceles with AB = AC. BA and CA are produced to Q and P such that AQ = AP. Show that PB = QC.

In ΔABP and ΔACQ:

AB = AC (Given)

AP = AQ (Given)

∠PAB = ∠QAC (Vertically opposite angles)

∴ ΔABP ≅ ΔACQ (By SAS congruence rule)

∴ PB = QC (By CPCT)

Question 9

In ΔABC, D is the midpoint of BC. DE ⊥ AB, DF ⊥ AC and DE = DF. Show that ΔBED ≅ ΔCFD.

In ΔBED and ΔCFD:

DE = DF (Given)

∠DEB = ∠DFC = 90° (DE ⊥ AB, DF ⊥ AC)

BD = CD (D is midpoint of BC)

∴ ΔBED ≅ ΔCFD (By RHS congruence rule)

Question 10

If the bisector of an angle of a triangle also bisects the opposite side, prove that the triangle is isosceles.

Let AD bisect ∠A in ΔABC where D is the midpoint of BC. Produce AD to E such that AD = DE. Join CE.

In ΔABD and ΔECD:

AD = ED (Construction)

BD = CD (D is midpoint — Given)

∠ADB = ∠EDC (Vertically opposite)

∴ ΔABD ≅ ΔECD (SAS) → AB = EC and ∠BAD = ∠CED (CPCT)

Since ∠BAD = ∠CAD (AD bisects ∠A) → ∠CAD = ∠CED → in ΔACE: AC = CE (sides opposite equal angles)

∴ AB = AC (both equal CE) → ΔABC is isosceles

Question 11

In right triangle ABC (right angle at B), ∠BCA = 2∠BAC. Show that hypotenuse AC = 2BC.

Produce CB to D such that BC = BD. Join AD.

In ΔABD and ΔABC:

BD = BC (Construction)

∠ABD = ∠ABC = 90°

AB = AB (Common)

∴ ΔABD ≅ ΔABC (SAS) → AD = AC and ∠BAD = ∠BAC (CPCT)

Let ∠BAC = x. Then ∠DAC = 2x. But ∠BCA = 2x (Given). In ΔADC: ∠DAC = ∠DCA = 2x → AD = DC. But AD = AC and DC = BD + BC = 2BC.

∴ AC = 2BC (Proved)

Which Congruence Rule Is Used — Quick Reference

Q1, Q2, Q4, Q5, Q7, Q8

Rule Applied

SAS — Side Angle Side

Two sides + included angle match

Rule Applied

AAS — Angle Angle Side

Two angles + one side match (vertically opp. angles + 90°)

Rule Applied

RHS — Right Hypotenuse Side

Right angle, hypotenuse, one side match

Q6, Q10, Q11

Rule Applied

SAS — with construction

Auxiliary point or line added to form congruent triangles

The most important skill in Exercise 7.1 is recognising which pair of triangles to work with and correctly identifying the matching elements. Always state the congruence rule explicitly, and then use CPCT to conclude about the required sides or angles.