Problem
1994 AHSME Problem 13
In triangle ABC, AB=AC. If there is a point P strictly between A and B such that AP=PC=CB, then \angle A = ![[asy] draw((0,0)--(8,0)--(4,12)--cycle); draw((8,0)--(1.6,4.8)); label("A", (4,12), N); label("B", (0,0), W); label("C", (8,0), E); label("P", (1.6,4.8), NW); dot((0,0)); dot((4,12)); dot((8,0)); dot((1.6,4.8)); [/asy]](/static/aops/302d3a9db01baaf4d628ba609f3c472775915aab.png)
\textbf{(A)}\ 30^{\circ} \qquad\textbf{(B)}\ 36^{\circ} \qquad\textbf{(C)}\ 48^{\circ} \qquad\textbf{(D)}\ 60^{\circ} \qquad\textbf{(E)}\ 72^{\circ}
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