Bikin A.d:

In triangle $ABC$, the midpoints of sides $AC$ and $BC$ are marked as points $M$ and $N$ respectively. Angle $MAN$ is $15^{\circ}$, and angle $BAN$ is $45^{\circ}$.

Find angle $ABM$.
Extend segment $M N$ by its length in both directions and obtain points $K$ and $L$ (see the left figure). Since $M$ is the common midpoint of segments $A C$ and $K N$, $A K C N$ is a parallelogram. Therefore, $\angle C K M=45^{\circ}, \angle K C M=15^{\circ}$. Mark a point $P$ on segment $C M$ such that $\angle C K P=15^{\circ}$. Then segment $K P$ will divide triangle $K C M$ into two isosceles triangles. Additionally, $\angle P M N=60^{\circ}$, so triangle $M P N$ is equilateral. Triangles $P L N$ and $P K M$ are congruent, triangle $C P L$ is isosceles and right-angled, $C L B M$ is a parallelogram, hence $\angle A B M=\angle C L N=\angle C L P+\angle M L P=75^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_476f7e28c93a861c2173g-44.jpg?height=1115&width=1604&top_left_y=11&top_left_x=227)

## Answer

$75^{\circ}$.