AHSME 1961 Problem 25

$\triangle ABC$ is isosceles with base $AC$. Points $P$ and $Q$ are respectively in $CB$ and $AB$ and such that $AC=AP=PQ=QB$. The number of degrees in $\angle B$ is: $\textbf{(A)}\ 25\frac{5}{7} \textbf{(B)}\ 26\frac{1}{3} \textbf{(C)}\ 30 \textbf{(D)}\ 40 \textbf{(E)}\ \text{Not determined by the information given}$