46. In a convex quadrilateral $A B C D$, point $M$ is the midpoint of side $A D$, $C M \| A B$, $A D=B D$ and $3 \angle B A C=\angle A C D$. Find the angle $A C B$.

(S. Berlov)
46. Answer: $\angle A C B=90^{\circ}$.

Notice that $\angle B A C=\angle A C M$, hence $\angle D C M=\angle A C D-\angle A C M=2 \angle A C M$. Let $N$ be the midpoint of segment $A B$. Then $D N$ is the median, bisector, and altitude of the isosceles triangle $A B D$. Next, point $C$ lies on the midline of this triangle, which is clearly the perpendicular bisector of $D N$, so $C D=C N$. Thus, triangle $N C D$ is isosceles and $C M$ is its bisector. Therefore, $\angle N C M=\angle D C M=2 \angle A C M$, hence $\angle N C A=\angle A C M=\angle N A C$. This means that triangle $A N C$ is isosceles and $N C=A N=N B$, but then

![](https://cdn.mathpix.com/cropped/2024_05_06_e84b7905ec5809a7a2b4g-2.jpg?height=351&width=312&top_left_y=133&top_left_x=1660)
$\angle A C B=90^{\circ}$.