Westwood's Puzzle/Proof 2

Proof
The crucial geometric truth to note is that:
 * $CJ = CG, AG = AF, BF = BJ$

This follows from the fact that:
 * $\triangle CEJ \cong \triangle CEG$, $\triangle AEF \cong \triangle AEG$ and $\triangle BEF \cong \triangle BEJ$

This is a direct consequence of the point $E$ being the center of the incircle of $\triangle ABC$.

Then it is just a matter of algebra.

Let $AF = a, FB = b, CJ = c$.