Westwood's Puzzle/Proof 2

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

Inscribe a circle $GFJ$ 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 $ABCD$.


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

\(\ds \paren {a + b}^2 + \paren {b + c}^2\) \(=\) \(\ds \paren {a + c}^2\) Pythagoras's Theorem
\(\ds \leadsto \ \ \) \(\ds a^2 + 2 a b + b^2 + b^2 + 2 b c + c^2\) \(=\) \(\ds a^2 + 2 a c + c^2\)
\(\ds \leadsto \ \ \) \(\ds a b + b^2 + b c\) \(=\) \(\ds a c\)
\(\ds \leadsto \ \ \) \(\ds a b + b^2 + b c + a c\) \(=\) \(\ds 2 a c\)
\(\ds \leadsto \ \ \) \(\ds \paren {a + b} \paren {b + c}\) \(=\) \(\ds 2 a c\)


Source of Name

This entry was named for Matt Westwood.