1965 AHSME Problems/Problem 30

Problem

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

[asy]  path circ, hyp;  draw((0,16)--(0,0)--(8,0)--(0,16)); dot((0,16)); label("A", (0,16), NW); dot((0,0)); label("C", (0,0), SW); dot((8,0)); label("B", (8,0), SE);  dot((0,8)); label("F",(0,8),W);  markscalefactor=0.1; draw(rightanglemark((0,16),(0,0),(8,0)));  circ=circle((4,0),4); draw(circ); hyp=(0,16)--(8,0); pair [] x=intersectionpoints(circ,hyp); dot(x[1]); label("D",x[1],NE); draw(x[1]--(0,8)); draw(x[1]--(0,0));  [/asy]

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 $\fbox{B}$.

See Also

1965 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 29
Followed by
Problem 31
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