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

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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)


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


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]


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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