• Prove that $BE\mid\mid CF$ and $BG\mid\mid CD$.
• Use ratio of lengths to finish the problem.

$BF=BC \implies \angle BCF = \angle BFC$

$Also, $\angle BCF = 180 - \angle BFC - \angle FBC $

$\implies BCF = \frac{180-\angle FBC}{2} = \frac{\angle ABC}{2} = \angle EBC$

So, $BE\mid\mid CF$

In the same way, $BG\mid\mid CD$

\[BE\mid\mid CF \implies \frac{AB}{AF} =\frac{AE}{AC}\]

\[BG\mid\mid CD \implies \frac{AB}{AD} =\frac{AG}{AC}\]

Dividing the equations

\[\frac{AF}{AD} = \frac{AG}{AE}\]

So, $GF\mid\mid DE$