In this chapter we will solve questions related to congruent triangles.

All the questions are to the standard of grade 09.

**Question 01**

In the below figure AD = BC. Prove that line CD bisects AB.

**Given:**

AD = BC

∠OAD = ∠OBC = 90 degree

**To prove:**

OA = OB

**Solution**

Consider triangle OBC and OAD

∠BOC = ∠DOA ( vertically opposite angle)

∠OBC = ∠OAD = 90 degree

BC = DA

By AAS congruency, both the triangles are congruent.

So OB = OA

Hence Proved

**Question 02**

In the below figure parallel lines l & m are intersected by parallel transversal p and q. Prove that triangle ABC and CDA are congruent.

**Solution**

Consider triangle ABC and CDA;

∠BAC = ∠ACD ( alternate interior angle )

AC = CA ( common side )

∠ACB = ∠DAC ( alternate interior angle )

By ASA congruency, both triangles are congruent.

Hence Proved.

**Question 03**

ABC is an isosceles triangle in which AB = AC. AD is the altitude. Prove that AD bisect BC.

**Given;**

AB = AC

∠ADB = 90 degree

∠B = ∠C (angle opposite to equal sides are equal)

**To prove;**

BD = DC

**Proof:**

Consider triangle ABD and ACD

AB = AC

AD = DA

∠ADB = ∠ADC = 90 degree

By RHS congruency, both triangles are congruent.

So, BD = DC

Hence Proved

**Question 04**

ABC & DBC are isosceles triangle.

Prove that;

(i) triangle ABD & ACD are congruent

(ii) triangle ABE & ACE are congruent

**Given:**

AB = AC

DB = DC

**(i) Prove triangle ABD & ACD are congruent **

Consider triangle ABD & ACD;

AB = AC ( given )

DB = DC ( given )

AD = DA ( common side )

By SSS congruency condition, both the triangles are congruent.

**(ii) Prove that triangle ABE & ACE are congruent **

Consider triangle ABE & ACE;

AB = AC

∠ABE = ∠ACE ( angle opposite to equal sides are equal)

AE = EA (common side)

By SAS congruency, both the triangles are congruent.

**Question 05**

In the below figure AB = CB and ∠x = ∠y. Prove that AE = CD

**Solution**

∠ADC = x

AB is a straight line and we know that sum of adjacent angle in straight line measure 180 degree.

∠BDC + ∠x = 180

∠BDC = 180 – ∠x

Similarly we can write the following;

∠ AEB + ∠y = 180

∠AEB = 180 – ∠y

Since ∠x =** ∠**y, we can say that ∠BDC = ∠ AEB

**Now consider triangle ABE and CBD;**

AB = CB ( given )

∠B = ∠B (common angle)

∠AEB = ∠BDC (proved above)

By AAS congruency, both triangles are congruent.

So, AE = CD

Hence Proved.

**Question 06**

In the below figure, line l bisect ∠A into two equal parts. Prove that triangle AQB & APB are congruent.

**Solution:**

Consider triangle AQB and APB;

∠QAB = ∠PAB ( line l is angle bisector)

∠ AQB = ∠ APB = 90 degree

AB = BA (common side)

By AAS congruency, both triangles are congruent.

Hence Proved.