Westwood's Puzzle/Proof 2
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Theorem
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 $ABCD$.
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 $AF = a, FB = b, CJ = c$.
\(\ds \paren {a + b}^2 + \paren {b + c}^2\) | \(=\) | \(\ds \paren {a + c}^2\) | Pythagoras's Theorem | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds a^2 + 2 a b + b^2 + b^2 + 2 b c + c^2\) | \(=\) | \(\ds a^2 + 2 a c + c^2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds a b + b^2 + b c\) | \(=\) | \(\ds a c\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds a b + b^2 + b c + a c\) | \(=\) | \(\ds 2 a c\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {a + b} \paren {b + c}\) | \(=\) | \(\ds 2 a c\) |
$\blacksquare$
Source of Name
This entry was named for Matt Westwood.