1975 AMC 12 Question 26 - thichhoctoan.com

Question 26

In acute triangle $ABC$ the bisector of $\angle A$ 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

$\begin{aligned}\angle CND+\angle BMD\\{}-\angle
DAC=120^\circ\end{aligned}$
$AMDN$ is a trapezoid
$BC$ is parallel to $MN$
$AM-AN=\frac{3(DB-DC)}2$
$AB-AC=\frac{3(DB-DC)}2$