Westwood's Puzzle

Theorem


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 $$AC$$ $$L$$.



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

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