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 $\fbox{B}$ .

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1965 AHSME Problems/Problem 30

Problem 30

Solution 1

Solution 2