9.20. (SFRY, 83). Inside triangle $A B C$, a point $M$ is taken, for which $\angle M B A=30^{\circ}, \angle M A B=10^{\circ}$. Find $\angle A M C$, if $\angle A C B=80^{\circ}$ and $A C=B C$.
9.20. Let the height $CH$ of triangle $ABC$ intersect the line

![](https://cdn.mathpix.com/cropped/2024_05_21_03a501298610b72904feg-183.jpg?height=490&width=690&top_left_y=1251&top_left_x=194)

Fig. 16
$BM$ at point $E$ (Fig. 16). Then $AE = BE$ and

\[
\begin{aligned}
& \angle EAM = \angle EAB - \angle MAB = 30^\circ - 10^\circ = 20^\circ, \\
& \angle ACE = (1/2) \angle ACB = 40^\circ, \\
& \angle EAC = \angle CAH - \angle EAB = (90^\circ - 40^\circ) - 30^\circ = 20^\circ, \\
& \angle AME = \angle MAB + \angle MBA = 10^\circ + 30^\circ = 40^\circ
\end{aligned}
\]

which means that triangles $AME$ and $ACE$ are equal by the common side and two angles. Therefore,

\[
AM = AC, \angle AMC = \angle ACM = (1/2)(180^\circ - \angle CAM) = 70^\circ
\]