![](https://cdn.mathpix.com/cropped/2024_05_06_476f7e28c93a861c2173g-32.jpg?height=292&width=2041&top_left_y=1432&top_left_x=0)

In triangle $ABC$, the angles are known: $\angle A=45^{\circ}, \angle B=15^{\circ}$. On the extension of side $AC$ beyond point $C$, point $M$ is taken such that $CM=2AC$. Find $\angle AMB$.
Mark off on segment $C B$ a segment $C K$ equal to segment $A C$.

## Solution

Let $P$ be the midpoint of segment $C M$. Then $\quad A C=C P=P M$. Mark a point $K$ on side $C B$ such that $C K=C A$. By the external angle theorem of a triangle,

$\angle P C K=\angle C A B+\angle A B C=45^{\circ}+15^{\circ}=60^{\circ}$.

Therefore, triangle $C P K$ is equilateral. Thus, $P C=P K=P M$. Consequently, triangle $C K M$ is right-angled and $\angle A M K=90^{\circ}-60^{\circ}=30^{\circ}$.

Since $\angle C A K=\angle C K A=30^{\circ}$, triangle $A K M$ is isosceles, and since $\angle K A B=\angle K B A=15^{\circ}$, triangle $A K B$ is also isosceles. Therefore, $K$ is the circumcenter of triangle $A M B$, and $\angle A M B=1 / 2 \angle A K B=75^{\circ}$.
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