# 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:

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

Similarly:

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

Finally:

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

$\blacksquare$

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

 $\displaystyle \paren {a + b}^2 + \paren {b + c}^2$ $=$ $\displaystyle \paren {a + c}^2$ Pythagoras's Theorem $\displaystyle \leadsto \ \$ $\displaystyle a^2 + 2 a b + b^2 + b^2 + 2 b c + c^2$ $=$ $\displaystyle a^2 + 2 a c + c^2$ $\displaystyle \leadsto \ \$ $\displaystyle a b + b^2 + b c$ $=$ $\displaystyle a c$ $\displaystyle \leadsto \ \$ $\displaystyle a b + b^2 + b c + a c$ $=$ $\displaystyle 2 a c$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {a + b} \paren {b + c}$ $=$ $\displaystyle 2 a c$

$\blacksquare$

## Source of Name

This entry was named for Matt Westwood.