Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

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

$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, $GF\mid\mid DE$