Westwood's Puzzle/Proof 1
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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
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 | |||||||||||
| \(\ds \) | \(=\) | \(\ds \Area \triangle HAL + \Area \triangle IKC + \Area \Box DHLKI\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \Area \triangle GEL + \Area \triangle GKE + \Area \Box DHLKI\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \Area \Box DHEI\) |
$\blacksquare$
Source of Name
This entry was named for Matt Westwood.