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 andΔDCB    

∠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 || BC and transversal AB cuts at A and B respectively 

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

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

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

Since AB and CD intersect at O.

∴∠AOD = ∠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