Westwood's Puzzle

Theorem

Take any rectangle $ABCD$ and draw the diagonal $AC$.

Inscribe a circle 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

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

Call the intersection of $IE$ and $AC$ $K$, and the intersection of $EH$ and $AC$ $L$.

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \angle CKI\)	\(=\)	\(\displaystyle \angle EKG\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By Two Straight Lines make Equal Opposite Angles
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \angle EGK\)	\(=\)	\(\displaystyle \mbox{Right Angle}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By Tangent to Circle is Perpendicular to Radius
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \angle KIC\)	\(=\)	\(\displaystyle \mbox{Right Angle}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Because $IF \perp CD$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \angle EGK\)	\(=\)	\(\displaystyle \angle KIC\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By Euclid's Fourth Postulate
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle IC\)	\(=\)	\(\displaystyle EJ\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By Parallel Lines are Everywhere Equidistant
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle EJ\)	\(=\)	\(\displaystyle EG\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Because both are radii of the same circle
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle IC\)	\(=\)	\(\displaystyle EG\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By Euclid's First Common Notion
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \mbox{Area}\triangle IKC\)	\(=\)	\(\displaystyle \mbox{Area}\triangle GKE\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By Triangle Angle-Angle-Side Equality
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \angle HLA\)	\(=\)	\(\displaystyle \angle GLE\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By Two Straight Lines make Equal Opposite Angles
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \angle EGL\)	\(=\)	\(\displaystyle \mbox{Right Angle}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By Tangent to Circle is Perpendicular to Radius
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \angle AHL\)	\(=\)	\(\displaystyle \mbox{Right Angle}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Because $HJ \perp AD$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \angle EGL\)	\(=\)	\(\displaystyle \angle AHL\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By Euclid's Fourth Postulate
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle HA\)	\(=\)	\(\displaystyle EF\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By Parallel Lines are Everywhere Equidistant
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle EF\)	\(=\)	\(\displaystyle EG\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Because both are radii of the same circle
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle HA\)	\(=\)	\(\displaystyle EG\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By Euclid's First Common Notion
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \mbox{Area}\triangle HAL\)	\(=\)	\(\displaystyle \mbox{Area}\triangle GEL\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By Triangle Angle-Angle-Side Equality
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \mbox{Area}\triangle ADC\)	\(=\)	\(\displaystyle \frac{AD\cdot CD} 2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By Area of a Triangle in Terms of Side and Altitude
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac{\mbox{Area}\Box ABCD} 2\)	\(=\)	\(\displaystyle \frac{AD\cdot CD} 2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By Area of a Parallelogram
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac{\mbox{Area}\Box ABCD} 2\)	\(=\)	\(\displaystyle \mbox{Area}\triangle ADC\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By Euclid's First Common Notion
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \mbox{Area}\triangle HAL + \mbox{Area}\triangle IKC + \mbox{Area}\Box DHLKI\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \mbox{Area}\triangle GEL + \mbox{Area}\triangle GKE+ \mbox{Area}\Box DHLKI\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \mbox{Area}\Box DHEI\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Alternative Proof

The crucial geometric truth to note is that (using the diagram from the first proof above) $CJ = CG, AG = AF, BF = BJ$.

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's just a matter of algebra.

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({a+b}\right)^2 + \left({b+c}\right)^2\)	\(=\)	\(\displaystyle \left({a+c}\right)^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Pythagoras's Theorem
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle a^2 + 2ab + b^2 + b^2 + 2bc + c^2\)	\(=\)	\(\displaystyle a^2 + 2ac + c^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle ab + b^2 + bc\)	\(=\)	\(\displaystyle ac\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle ab + b^2 + bc + ac\)	\(=\)	\(\displaystyle 2ac\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left({a + b}\right) \left({b + c}\right)\)	\(=\)	\(\displaystyle 2ac\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Source of Name

This entry was named for Matt Westwood.