#### Question

BD and CE are bisectors of ∠B and ∠C of an isosceles ΔABC with AB = AC. Prove that BD = CE.

#### Solution

Given that Δ*AB**C *is isosceles with *AB = AC *and *BD *and *CE *are bisectors of ∠*B *and ∠*C*

We have to prove *B**D *= *C**E *

Since *AB *= *AC ⇒* ∠A*BC *= ∠*ACB …….(1)*

* [∵ Angles opposite to equal sides are equal] *

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

*`∠ABD *= ∠*DBC *= ∠*BCE *= *ECA *=`(∠B)/2=(∠C)/2` …….(2)

Now,

Consider Δ*EB**C *andΔ*DCB *

*∠EBC *= ∠*DCB *[∵ ∠*B *= ∠*C *] from (1)

*B**C *= *B**C [Common side] *

*∠BCE *= ∠*C**BD [ ∵ From (2)] *

So, by ASA congruence criterion, we have *ΔEBC ≅Δ**D**CB *

Now,

*C**E *= *B**D [∵ Corresponding parts of congruent triangles are equal] *

or *B**D *= *CE*

∴Hence proved

Since *A**D *|| *BC *and transversal AB cuts at A and B respectively

∴∠*D**A**O *= ∠*O**B**C *……….(2) [alternate angle]

And similarly respectively *A**D *|| *BC *and transversal DC cuts at D ad C respectivaly

∴ ∠*A**D**O *= ∠*O**CB *………(3) [alternate angle]

Since AB and CD intersect at O.

∴∠*AOD *= ∠*BOC [Vertically opposite angles]*

Now consider *Δ**AOD *and *Δ**BOD*

*∠D**AO *= ∠*O**BC [∵ From (2)]*

*A**D *= *B**C [ ∵ From (1)]*

And *∠**ADO *= *∠**OCB [From (3)] *

So, by ASA congruence criterion, we have

*ΔA**O**D ≅**Δ**B**O**C *

Now,

*A**O *= *O**B *and *DO *= *OC [**∵*Corresponding parts of congruent triangles are equal]

⇒Lines AB and *C**D *bisect at O.

∴Hence proved