Question

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

Answer 1

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≅Δ DCB 

Now, 

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

or BD=CE

∴Hence proved 

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

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

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

∴ ∠ADO = ∠OCB        ………(3) [alternate angle] 

Since AB and CD intersect at O. 

∴ ∠ADO =∠BOC         [Vertically opposite angles] 

 Now consider ΔAOD and ΔBOD     

∠DAO=∠OBC             [∵ From (2)] 

AD = BC                   [ ∵ From (1)] 

And ∠ADO = ∠OCB         [From (3)] 

 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