Westwood's Puzzle/Proof 2

Theorem

 * WestwoodsPuzzle.png

Take any rectangle $ABCD$ and draw the diagonal $AC$.

Inscribe a circle in one of the resulting triangles $\triangle ABC$.

Drop perpendiculars $IEF$ and $HEJ$ from the center of this incircle $E$ to the sides of the rectangle.

Then the area of the rectangle $DHEI$ equals half the area of the rectangle $ABCD$.

Proof

 * Westwood's Puzzle Proof.png

The crucial geometric truth to note is that (using the diagram) $CJ = CG, AG = AF, BF = BJ$.

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's just a matter of algebra.

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