HMMT-Feb 2013 geo, Problem 6

Let $A B C D$ be a quadrilateral such that $\angle A B C=\angle C D A=90^{\circ}$, and $B C=7$. Let $E$ and $F$ be on $B D$ such that $A E$ and $C F$ are perpendicular to $B D$. Suppose that $B E=3$. Determine the product of the smallest and largest possible lengths of $D F$.

Solution: By inscribed angles, $\angle C D B=\angle C A B$, and $\angle A B D=\angle A C D$. By definition, $\angle A E B=\angle C \overrightarrow{D A}=\angle A B C=\angle C F A$. Thus, $\triangle A B E \sim \triangle A D C$ and $\triangle C D F \sim \triangle C A B$. This shows that \[ \frac{B E}{A B}=\frac{C D}{C A} \text { and } \frac{D F}{C D}=\frac{A B}{B D} \] Based on the previous two equations, it is sufficient to conclude that $3=E B=F D$. Thus, $F D$ must equal to 3 , and the product of its largest and smallest length is 9 . $\fbox{9}$.

Answer: 9