# Westwood's Puzzle

## Contents

## 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.