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**Q. (1) In fig. (10).40, it is given that RT = TS, ∠ 1 = 2 ∠ 2 and 4 = 2 ∠ (3) Prove that ΔRBT ≅ ΔSAT.**

**Solution:**

In the figure, given that

RT = TS ……(i)

∠ 1 = 2 ∠ 2 ……(ii)

And ∠ 4 = 2 ∠ 3 ……(iii)

To prove that ΔRBT ≅ ΔSAT.

Let the point of intersection RB and SA be denoted by O

Since RB and SA intersect at O

∠ AOR = ∠ BOS [Vertically opposite angles]

- ∠ 1 = ∠ 4
- 2 ∠ 2 = 2 ∠ 3 [From (ii) and (iii)]
- ∠ 2 = ∠ 3 ……(iv)

Now we have RT =TS in Δ TRS

Δ TRS is an isosceles triangle

∠ TRS = ∠ TSR ……(v)

But we have

∠ TRS = ∠ TRB + ∠ 2 ……(vi)

∠ TSR = ∠ TSA + ∠ 3 ……(vii)

Putting (vi) and (vii) in (v) we get

∠ TRB + ∠ 2 = ∠ TSA + ∠ B

=> ∠ TRB = ∠ TSA [From (iv)]

Now consider Δ RBT and Δ SAT

RT = ST [From (i)]

∠ TRB = ∠ TSA [From (iv)] ∠ RTB = ∠ STA [Common angle]

From ASA criterion of congruence, we have

Δ RBT = Δ SAT

** **

**Q. (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 that lines AB and CD Intersect at O

** **

Such that BC ∥ AD and BC = AD …….(i)

We have to prove that AB and CD bisect at O.

To prove this first we have to prove that Δ AOD ≅ Δ BOC

** **

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

**Solution:**

Given that Δ ABC is isosceles with AB = AC and BD and CE are bisectors of ∠ B and ∠ C We have to prove BD = CE

Since AB = AC

=> Δ ABC = Δ ACB ……(i)

[Angles opposite to equal sides are equal]

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

- ∠ ABD = ∠ DBC = ∠ BCE = ECA =

Now,

Consider Δ EBC = Δ DCB

∠ EBC = ∠ DCB [∠ B = ∠ C] [From (i)]

BC = BC [Common side]

∠ BCE = ∠ CBD [From (ii)]

So, by ASA congruence criterion, we have Δ EBC ≅ Δ DCB

Now,

CE = BD [Corresponding parts of congruent triangles we equal]

or, BD = CE

Hence proved

Since AD ∥ BC and transversal AB cuts at A and B respectively

∠ DAO = ∠ OBC …….(ii) [alternate angle]

And similarly AD ∥ BC and transversal DC cuts at D and C respectively

∠ ADO = ∠ OBC ..…(iii) [alternate angle]

Since AB end CD intersect at O.

∠ AOD = ∠ BOC [Vertically opposite angles]

Now consider Δ AOD and Δ BOD

∠ DAO = ∠ OBC [From (ii)]

AD = BC [From (i)]

And ∠ ADO = ∠ OCB [From (iii)]

So, by ASA congruence criterion, we have

ΔAOD≅ΔBOC

Now,

AO= OB and DO = OC [Corresponding parts of congruent triangles are equal)

- Lines AB and CD bisect at O.

Hence proved

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