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

All the questions are to the standard of grade 09 math.

**Question 01**

Given below is the triangle in which TR = TS. Also ∠1 = 2 (∠2) and ∠4 = 2 (∠3). Prove that triangle RBT and SAT are congruent.

**Solution**

∠1 = ∠4 (vertically opposite angles)

2 (∠2) = 2 (∠3)

∠2 = ∠3

In triangle angle opposite to equal sides are equal.

TR = TS

∠R = ∠S

∠TRB + ∠2 = ∠TSA + ∠3

Since ∠2 & ∠3 are equal, we can cancel each of them from the equation.

∠TRB = ∠TSA

Now consider triangle RBT and SAT

∠TRB = ∠TSA

RT = ST (given )

∠T = ∠T (common angle)

By ASA congruency, both the triangles are congruent.

Hence Proved.

**Question 02**

In the below triangle AB = AC. Also BD & CE are bisectors of respective angles. Prove that BD = CE.

**Solution**

Since BD & CE are angle bisectors, we can write;

∠1 = ∠2

∠3 = ∠4

In triangle, angle opposite to equal sides are equal.

∠B = ∠C

∠1 + ∠2 = ∠3 + ∠4

2 (∠1) = 2 (∠3)

∠1 = ∠3

Now consider triangle BDC and CEB

∠C = ∠B

BC = CB ( common side)

∠1 = ∠3

By AAS congruency, both triangles are congruent.

So, BD = CE

Hence Proved

**Question 03**

In the below image PQ = PR. Line ST is parallel to QR. Prove that PS = PT

**Solution**

We know that in triangle angle opposite to equal sides are equal.

Since PQ = PR, we can write ∠Q = ∠R

ST & QR are parallel lines intersected by transversal PQ & PR.

∠S = ∠Q (corresponding angles)

∠T = ∠R (corresponding angles)

Combining all the expression we get;

∠S = ∠T

In triangle, sides opposite to equal angles are equal.

Since ∠S = ∠T, we can write PS = PT

Hence Proved.

**Question 04**

In the below figure AB = CD and AD = CB. Prove that triangle ADC & CBA are congruent.

**Solution**

Considering triangle ADC and CBA.

AD = CB

CD = BA

CA = AC

By SSS congruency, both the triangles are congruent.

**Question 05**

In the below triangle, D is the midpoint of BC. DP & DQ are equal perpendicular line on sides AB and AC respectively. Prove that AB = AC.

**Solution**

Consider triangle BPD and CQD

BD = DC

PD = DQ

∠DPB = ∠DQC

By RHS congruency, both the triangles are congruent.

So BP = CQ

Now consider triangle APD & AQD

DP = DQ

∠APD = ∠AQD = 90

AD = DA

By RHS congruency, both triangles are congruent.

So, AP = AQ

Adding PB on both sides;

AP + PB = AQ + PB

AB = AQ + QC

AB = AC

Hence, given figure is an isosceles triangle.

**Question 06**

In below figure BE and CE are equal perpendicular lines. Prove that the triangle is isosceles.

**Solution**

Consider triangle BFC & CEF.

CF = BE (given)

∠CFB = ∠BEC = 90

BC = CB ( common side)

By RHS congruency, both triangles ae congruent.

So, ∠FBC = ∠ECB

we know that in triangle, side opposite to equal angle are equal.

Since ∠B = ∠C, we can write AB = AC

Hence, the triangle is isosceles.