Question
$\triangle ABC$ in a isosceles triangle with $AB = AC. D$ is a point on $BC$ produced. $ED$ intersects $AB$ at $E$ and $AC$ at $F.$ Prove that $AF > AE.$
✓
Answer
$\angle AEF > \angle ABC ...($Exterior angle property$)$
$\angle AEF = \angle DFC$
$\angle ACB > \angle DFC ...($Exterior angle property$)$
$\Rightarrow \angle ACB > \angle AFE$
Since $AB = AC$
$\Rightarrow \angle ACB = \angle ABC$
So, $\angle ABC > \angle AFE$
$\Rightarrow \angle AEF > \angle ABC > \angle AFE$
that is $\angle AEF > \angle AFE$
$\Rightarrow AF > AE.$
$\angle AEF = \angle DFC$
$\angle ACB > \angle DFC ...($Exterior angle property$)$
$\Rightarrow \angle ACB > \angle AFE$
Since $AB = AC$
$\Rightarrow \angle ACB = \angle ABC$
So, $\angle ABC > \angle AFE$
$\Rightarrow \angle AEF > \angle ABC > \angle AFE$
that is $\angle AEF > \angle AFE$
$\Rightarrow AF > AE.$
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