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

We have to prove BD = CE

Since AB = AC ⇒ ∠ABC=∠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 ΔEBC and ΔDCB

∠EBC=∠DCB [∵ ∠B = ∠C] from (1)

BC = BC [Common side]

∠BCE = ∠CBD [ ∵ From (2)]

So, by ASA congruence criterion, we have ΔEBC ≅ ΔDCB

Now,

CE = BD [∵ Corresponding parts of congruent triangles are equal]

or, BD = CE

∴ Hence proved

Since AD|| BCand transversal AB cuts at A and B respectively

∠DAO = ∠OBC ......(2) [alternate angle]

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

∠ADO = ∠OCB ......(3) [alternate angle]

Since AB and CD intersect at O.

∠AOD = ∠BOC [Vertically opposite angles]

Now,

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

⇒ Lines AB and CD bisect at O.

∴ Hence proved