Question

In the given figure if $AB=AC$, prove that $AF>AE$

Answer 1

Solution:-

In $△ABC$,

$AB=AC\left[\text{Given}\right]$

$\therefore \angle ACB=\angle ABC.....\left(i\right)[\because \text{Angle opposite to equal sides are equal}]$

In $△CDF$,

$\angle ACB$ is an exterior angle.

$\therefore \angle ACB>\angle DFC[\because \text{Exterior angle property}]$

$\Rightarrow \angle ABC>\angle DFC.....\left(ii\right)\left[\text{From}\left(i\right)\right]$

Also,

$\angle DFC=\angle AFE\left[\text{Vertically opposite angles}\right]$

$\Rightarrow \angle ABC>\angle AFE.....\left(iii\right)\left[\text{From}\left(ii\right)\right]$

In $△BDE,\angle DEA$ is an exterior angle

$\therefore \angle DEA>\angle ABC$

$\Rightarrow \angle DEA>\angle AFE\left[\text{From}\left(iii\right)\right]$

In $△AEF$,

$\angle AEF>\angle AFE$

$\therefore AF>AE[\because \text{Side opposite to greater angle is greater}]$

Hence proved.