1975 AHSME Problems/Problem 24

Problem

In triangle $ABC$ , $\angle C = \theta$ and $\angle B = 2\theta$ , where $0^{\circ} < \theta < 60^{\circ}$ . The circle with center $A$ and radius $AB$ intersects $AC$ at $D$ and intersects $BC$ , extended if necessary, at $B$ and at $E$ ( $E$ may coincide with $B$ ). Then $EC = AD$

$\textbf{(A)}\ \text{for no values of}\ \theta \qquad \textbf{(B)}\ \text{only if}\ \theta = 45^{\circ} \\ \textbf{(C)}\ \text{only if}\ 0^{\circ} < \theta \leq 45^{\circ} \qquad \textbf{(D)}\ \text{only if}\ 45^{\circ} \leq \theta \leq 60^{\circ} \\ \textbf{(E)}\ \text{for all}\ \theta\ \text{such that}\ 0^{\circ} < \theta < 60^{\circ}$

Solution

Since $AD = AE$ , we know $EC = AD$ if and only if triangle $ACE$ is isosceles and $\angle ACE = \angle CAE$ . Letting $\angle ACE = \theta$ , we want to find when $\angle CAE = \theta$ . We know $\angle ABC = \angle AEB = 2\theta$ , so $\angle EAB = 180-4\theta$ . We also know $\angle CAB = 180-3\theta$ , and since $\angle CAE = \angle CAB - \angle EAB$ , $\angle CAE = \theta$ . Since we now know that $\angle CAE = \angle ACE$ regardless of $\theta$ , we have $0^{\circ} < \theta < 60^{\circ}$ , or $\boxed{E}$ .

1975 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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1975 AHSME Problems/Problem 24

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