# Class 9 RD Sharma Solutions – Chapter 10 Congruent Triangles- Exercise 10.2

**Question 1: In figure, it is given that RT = TS, ∠1 = 2∠2 and ∠4 = 2(∠3). Prove that **ΔRBT ≅ ΔSAT**.**

**Solution: **

In the figure,

Given:

RT = TS —-> (equation 1)

∠1 = 2 ∠2 —-> (equation 2)

∠4 = 2 ∠3 —->(equation 3)

To prove:

ΔRBT ≅ ΔSATLet the point of intersection RB and SA be denoted by O

∠AOR = ∠BOS [Vertically opposite angles]

So, ∠1 = ∠4

2 ∠2 = 2 ∠3 [From (equation 2) and (equation 3)]

or

∠2 = ∠3 —->(equation 4)Now in Δ TRS,

We have RT = TS

Δ TRS is an isosceles triangle

∠TRS = ∠TSR —->(equation 5)

But, ∠TRS = ∠TRB + ∠2 —->(equation 6)

∠TSR = ∠TSA + ∠3 —->(equation 7)

Putting (equation 6) and (equation 7) in (equation 5) we get

∠TRB + ∠2 = ∠TSA + ∠3

∠TRB = ∠TSA [From (equation 6)]

Consider ΔRBT and ΔSAT

RT = ST [From (equation 1)]

∠TRB = ∠TSA [From (equation 6)]

By ASA criterion of congruence, we have

ΔRBT ≅ ΔSAT

**Question 2: Two lines AB and CD intersect at O such that BC is equal and parallel to AD. Prove that the lines AB and CD bisect at O.**

**Solution: **

Given:

Lines AB and CD Intersect at O

Such that BC ∥ AD and

BC = AD —->(equation 1)

To prove:

AB and CD bisect at O.First we have to prove that ΔAOD ≅ ΔBOC

∠OCB =∠ODA {AD||BC and CD is transversal}

AD = BC {from (equation 1)}

∠OBC = ∠OAD {AD||BC and AB is transversal}

By ASA Criterion:

ΔAOD ≅ ΔBOC

OA = OB and OD = OC (By Corresponding parts of congruent triangle )

Therefore,

AB and CD bisect each other at O.

Hence, Proved.

**Question 3: BD and CE are bisectors of ∠B and ∠C of an isosceles **Δ** ABC with AB = AC. Prove that BD = CE.**

**Solution:**

Given:

ΔABC is isosceles

Where,

AB = AC

BD and CE are bisectors of ∠ B and ∠ C

To prove:

BD = CESince AB = AC (given)

∠ABC = ∠ACB —->(equation 1) {Angles opposite to equal sides are equal}

Since BD and CE are bisectors of ∠ B and ∠ C

∠ABD = ∠DBC =∠ BCE =ECA =∠B=∠C—-> (equation 2)

2 2

Now, Consider ΔEBC = ΔDCB

∠EBC = ∠DCB {From (equation 1)}

BC = BC {Common side}

∠BCE = ∠CBD {From (equation 2)}

By ASA congruence criterion,

Δ EBC ≅ Δ DCB

Since corresponding parts of congruent triangles are equal.(c.p.c.t.)

CE = BD

or,

BD = CE

Hence, proved.