Question

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

Answer 1

In $△ABC$

$AB=AC$ [Given]

$\angle ABC=\angle ACB......\left(1\right)$ [Angles opposite to equal sides of a trinagle are equal]

Also, $\frac{1}{2}AB=\frac{1}{2}AC$

$BE=CD.......\left(2\right)$ [Halves of equals are equal]

Since BD and CE are two medians

Now,

In $△BDC$ and $△CEB$

From (1) $\angle BCD=\angle CBE$

From (2) $BE=CD$

$BC=CB$ [Common]

$\therefore △BDC\cong △CEB$ [SAS Congruence rule]

$\therefore BD=CE$ [By CPCT]