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
