Question

In the given figure, $AD=AE$. $D$ and $E$ are points on $BC$, such that $BD=EC$. Prove that $AB=AC$.

Answer 1

Given that: $AD=AE,BD=EC$

To prove: $AB=AC$

Since, $AD=AE$ $△ADE$ is an isosceles triangle.

$AF\perp DE,DF=FE$ (Properties of isosceles triangle)

Now, $BD+DF=FE+EC\Rightarrow BF=FC$

In $△AFB$ and $△AFC$,

$\Rightarrow BF=FC$

$\angle AFB=\angle AFC$

$\Rightarrow AF$ is common side

Thus $△AFB\cong AFC$

Therefore, $AB=AC$ .....(By CPCT)