3. Given an isosceles triangle $ABC$ where $CA = CB$ and $\measuredangle ACB = 80^\circ$. Let $M$ be a point inside triangle $ABC$ such that $\measuredangle MBA = 30^\circ$ and $\measuredangle MAB = 10^\circ$. Determine $\measuredangle AMC$.
Solution. Let $N$ be the intersection of line $BM$ and the altitude from vertex $C$ of triangle $ABC$, as shown in the diagram on the right. Triangle $ABN$ is isosceles with base $AB$ and the angle at the vertex

$$
\measuredangle ANB=180^{\circ}-2 \cdot 30^{\circ}=120^{\circ}
$$

Furthermore, $\angle ANC=\measuredangle BNC=120^{\circ}$. Since $AN=AN, \measuredangle ACN=\measuredangle AMN=40^{\circ}$ and $\measuredangle ANC=\measuredangle ANM$, we conclude that triangles $ANC$ and $ANM$ are congruent. Therefore,

![](https://cdn.mathpix.com/cropped/2024_06_05_f7eeff4e7a5fefb59aecg-2.jpg?height=438&width=528&top_left_y=252&top_left_x=957)
$AC=AM$, i.e., triangle $ACM$ is isosceles with base $CM$. Finally,

$$
\measuredangle AMC=\frac{1}{2}\left(180^{\circ}-\measuredangle CAM\right)=\frac{1}{2}\left(180^{\circ}-40^{\circ}\right)=70^{\circ}
$$