Track / Stage 3 / 63 of 260 #63 of 1964

Problem 63

AMC 10/12, early questions

Geometry Difficulty 3.2 Multiple choice

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

Pick one

Official solution

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 $\fbox{B}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.