Westwood's Puzzle

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Theorem

WestwoodsPuzzle.png

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 1

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

Let $K$ be the intersection of $IE$ and $AC$.

Let $L$ be the intersection of $EH$ and $AC$.

Westwood's Puzzle Proof.png


First we have:

\(\text {(1)}: \quad\) \(\ds \angle CKI\) \(=\) \(\ds \angle EKG\) Two Straight Lines make Equal Opposite Angles
\(\ds \angle EGK\) \(=\) \(\ds \text {Right Angle}\) Tangent to Circle is Perpendicular to Radius
\(\ds \angle KIC\) \(=\) \(\ds \text {Right Angle}\) as $IF \perp CD$
\(\text {(2)}: \quad\) \(\ds \therefore \ \ \) \(\ds \angle EGK\) \(=\) \(\ds \angle KIC\) Euclid's Fourth Postulate
\(\ds IC\) \(=\) \(\ds EJ\) Opposite Sides and Angles of Parallelogram are Equal
\(\ds EJ\) \(=\) \(\ds EG\) as both are radii of the same circle
\(\text {(3)}: \quad\) \(\ds \therefore \ \ \) \(\ds IC\) \(=\) \(\ds EG\) Euclid's First Common Notion
\(\ds \therefore \ \ \) \(\ds \Area \triangle IKC\) \(=\) \(\ds \Area \triangle GKE\) Triangle Angle-Angle-Side Equality: $(1)$, $(2)$ and $(3)$


Similarly:

\(\text {(4)}: \quad\) \(\ds \angle HLA\) \(=\) \(\ds \angle GLE\) Two Straight Lines make Equal Opposite Angles
\(\ds \angle EGL\) \(=\) \(\ds \text {Right Angle}\) Tangent to Circle is Perpendicular to Radius
\(\ds \angle AHL\) \(=\) \(\ds \text {Right Angle}\) as $HJ \perp AD$
\(\text {(5)}: \quad\) \(\ds \therefore \ \ \) \(\ds \angle EGL\) \(=\) \(\ds \angle AHL\) Euclid's Fourth Postulate
\(\ds HA\) \(=\) \(\ds EF\) Opposite Sides and Angles of Parallelogram are Equal
\(\ds EF\) \(=\) \(\ds EG\) as both are radii of the same circle
\(\text {(6)}: \quad\) \(\ds \therefore \ \ \) \(\ds HA\) \(=\) \(\ds EG\) Euclid's First Common Notion
\(\ds \therefore \ \ \) \(\ds \Area \triangle HAL\) \(=\) \(\ds \Area \triangle GEL\) Triangle Angle-Angle-Side Equality: $(4)$, $(5)$ and $(6)$


Finally:

\(\ds \frac {\Area \Box ABCD} 2\) \(=\) \(\ds \frac {AD \cdot CD} 2\) Area of Parallelogram
\(\ds \) \(=\) \(\ds \Area \triangle ADC\) Area of Triangle in Terms of Side and Altitude
\(\ds \) \(=\) \(\ds \Area \triangle HAL + \Area \triangle IKC + \Area \Box DHLKI\)
\(\ds \) \(=\) \(\ds \Area \triangle GEL + \Area \triangle GKE + \Area \Box DHLKI\)
\(\ds \) \(=\) \(\ds \Area \Box DHEI\)

$\blacksquare$


Proof 2

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.