BD and CE are bisectors of ∠B and ∠C of an isosceles ΔABC with AB = AC. Prove that BD = CE
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
⟹ ΔABC = ΔACB ... (i)
[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
Now,
Consider ΔEBC = ΔDCB
∠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 ΔAOD and ΔBOD
∠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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