# 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 $\boxed{E}$.