# Westwood's Puzzle

## 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 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 $AC$.

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 \ \$ $\ds IC$ $=$ $\ds EG$ Common Notion $1$ $\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 \ \$ $\ds HA$ $=$ $\ds EG$ Common Notion $1$ $\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.