# 1965 AHSME Problems/Problem 30

## Problem 30

Let $BC$ of right triangle $ABC$ be the diameter of a circle intersecting hypotenuse $AB$ in $D$. At $D$ a tangent is drawn cutting leg $CA$ in $F$. This information is not sufficient to prove that

$\textbf{(A)}\ DF \text{ bisects }CA \qquad \textbf{(B) }\ DF \text{ bisects }\angle CDA \\ \textbf{(C) }\ DF = FA \qquad \textbf{(D) }\ \angle A = \angle BCD \qquad \textbf{(E) }\ \angle CFD = 2\angle A$

## Solution 1

We will prove every result except for $\fbox{B}$.

By Thales' Theorem, $\angle CDB=90^\circ$ and so $\angle CDA= 90^\circ$. $FC$ and $FD$ are both tangents to the same circle, and hence equal. Let $\angle CFD=\alpha$. Then $\angle FDC = \frac{180^\circ - \alpha}{2}$, and so $\angle FDA = \frac{\alpha}{2}$. We also have $\angle AFD = 180^\circ - \alpha$, which implies $\angle FAD=\frac{\alpha}{2}$. This means that $CF=DF=FA$, so $DF$ indeed bisects $CA$. We also know that $\angle BCD=90-\frac{180^\circ - \alpha}{2}=\frac{\alpha}{2}$, hence $\angle A = \angle BCD$. And $\angle CFD=2\angle A$ as $\alpha = \frac{\alpha}{2}\times 2$.

Since all of the results except for $B$ are true, our answer is $\fbox{B}$.

## Solution 2

It's easy to verify that $\angle CDA$ always equals $90^\circ$. Since $\angle CDF$ changes depending on the sidelengths of the triangle, we cannot be certain that $\angle CDF=45^\circ$. Hence our answer is $\fbox{B}$.