Question

$BD$ and $CE$ are bisectors of $\angle B$ and $\angle C$ of an isosceles $△ABC$ with $AB=AC.$ Prove that $BD=CE.$

Answer 1

Solution:
Given: $AB=AC$ and $BD$ and $CE$ are the bisectors of $\angle B$ and $\angle C$ respectively
To prove : $BD=CE$
Proof : In $△ABC,$
$AB=AC$
$\Rightarrow \angle B=\angle C\dots \left(i\right)$
$\text{(Angles opposite to equal sides are equal)}$
$\Rightarrow \frac{1}{2}\angle B=1/2\angle C$
$\Rightarrow \angle DBC=\angle ECB\dots \left(ii\right)$



Now, in $△DBC$ and $△EBC,$
$BC=BC\text{(Common)}$
$\angle C=\angle B\text{[ from}\left(i\right)\text{]}$
$\angle DBC=\angle ECB\text{[ from}\left(ii\right)\text{]}$
$\therefore △DBC\cong EBC\text{(by ASA axiom)}$

Thus, $BD=CE\text{(by C.P.C.T)}$
Hence, proved.