Westwood's Puzzle

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
Construct the perpendicular from $E$ to $AC$, and call its foot $G$.

Call the intersection of $IE$ and $AC$ $K$, and the intersection of $EH$ and $AC$ $L$.


 * Westwood's Puzzle Proof.png

Alternative Proof
The crucial geometric truth to note is that (using the diagram from the first proof above) $CJ = CG, AG = AF, BF = BJ$.

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