AHSME 1975 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 $\textbf{(A)}\ \measuredangle CND+\measuredangle BMD-\measuredangle DAC =120^{\circ} \textbf{(B)}\ AMDN\ \text{is a trapezoid} \textbf{(C)}\ BC\ \text{is parallel to}\ MN\\ \textbf{(D)}\ AM-AN=\frac{3(DB-DC)}{2} \textbf{(E)}\ AB-AC=\frac{3(DB-DC)}{2}$