# A Property of Right Trapezoids

The following statement has been communicated by Sohail Farhangi:

Given right trapezoid $ABCD$ with right angles at $A$ and $B$. Let circle $C(M)$ with $CD$ as diameter intersect $AB$ in $F$ and $H$. Then the perpendicular from $A$ to $DF$ and the perpendicular from $B$ to $CF$ meet on $CD$.

The applet below serves a dynamic illustration

18 January 2015, Created with GeoGebra

Created with GeoGebra

Given right trapezoid $ABCD$ with right angles at $A$ and $B$. Let circle $C(M)$ with $CD$ as diameter intersect $AB$ in $F$ and $H$. Then the perpendicular from $A$ to $DF$ and the perpendicular from $B$ to $CF$ meet on $CD$.

### Proof

Let $K$ and $G$ denote the intersections of the perpendicular from $B$ to $CF$ with $CF$ and $CD$, respectively. Since $\angle CFD$ subtends in $C(M)$ diameter $CD$, it is right: $\angle CFD=90^{\circ}$. This makes triangles $CDF$ and $CGK$ similar, implying the proportion $\displaystyle\frac{CG}{DG}=\frac{CK}{FK}$. Noting that the right triangles $BCK$ and $FBK$ are similar, we may continue

\displaystyle \begin{align} \frac{CG}{DG} &= \frac{CK}{FK} \\ &=\frac{CK}{BK}\times\frac{BK}{FK} \\ &= \mbox{tan}\angle CBK\space\times\space\mbox{tan}\angle BFK \\ &= (\mbox{tan}\angle BFC)^{2}. \end{align}

For the perpendicular from $A$, assume it intersects $DF$ in $J$ and $CD$ in $G'$. Then, similarly to the above,

\displaystyle \begin{align} \frac{CG'}{DG'} &= \frac{FJ}{DJ} \\ &=\frac{FJ}{AJ}\times\frac{AJ}{DJ} \\ &= \mbox{tan}\angle FAJ\space\times\space\mbox{tan}\angle ADJ \\ &= (\mbox{tan}\angle ADF)^{2}. \end{align}

But, since angles $BFC$ and $AFD$ are complementary, $\angle BFC=\angle ADF$ and, therefore, $\displaystyle\frac{CG}{DG}=\frac{CG'}{DG'}$ such that $G=G'$, and we are done.

This configuration has additional properties. For examples, $\angle AGB$ is right and $HG\perp CD$ so that quadrilaterals $BCGH$ and $ADGH$ are both cyclic.

Finally, circle $C(M)$ place an ancilary role, producing point $F$ on $AB$ such that $\angle CFD=90^{\circ}$. Thus given such point a priori, we could procede with the construction as shown below:

### Thébault's Problems

• Thébault's Problem I
• Thébault's Problem II
• Thébault's Problem III
• Y. Sawayama's Lemma
• Jack D'Aurizio Proof of Sawayama's Lemma
• Y. Sawayama's Theorem
• Thébault's Problem III, Proof (J.-L. Ayme)
• Circles Tangent to Circumcircle
• Thébault's Problem IV
• A Lemma on the Road to Sawayama
• Excircles Variant of Thébault's Problem III
• In the Spirit of Thebault I
• Dao's Variant of Thebault's First Problem