1961 AHSME 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}\qquad \textbf{(B)}\ 26\frac{1}{3}\qquad \textbf{(C)}\ 30\qquad \textbf{(D)}\ 40\qquad \textbf{(E)}\ \text{Not determined by the information given}

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.