# Congruent triangle problems – set 02

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

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

By RHS congruency, both triangles are congruent.
So, AP = AQ

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.