1975 AHSME Problem 26

In acute \triangle ABC the bisector of \measuredangle A meets side BC at D. The circle with center B and radius BD intersects side AB at M; and the circle with center C and radius CD intersects side AC at N. Then it is always true that

\begin{array}{l} \textbf{(A)}\ \measuredangle CND+\measuredangle BMD-\measuredangle DAC =120^{\circ} \qquad \textbf{(B)}\ AMDN\ \text{is a trapezoid}\qquad \textbf{(C)}\ BC\ \text{is parallel to}\ MN\\ \qquad \textbf{(D)}\ AM-AN=\frac{3(DB-DC)}{2}\qquad \textbf{(E)}\ AB-AC=\frac{3(DB-DC)}{2} \end{array}

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