[ The sum of the angles of a triangle. The theorem about the exterior angle.] [ Criteria and properties of an isosceles triangle. ]

On the side $B C$ of the isosceles triangle $A B C(A B=B C)$, points $N$ and $M$ ( $N$ is closer to $B$ than $M$) are taken such that $N M=A M$ and $\angle M A C=\angle B A N$.

Find $\angle C A N$.
Let $\angle M A C=\angle B A N=\alpha, \angle N A M=\angle A N M=\varphi$. Since $A M C$ is the exterior angle of triangle $A M N$, then $\angle A M C=2 \varphi . \angle A C M=\angle C A B=2 \alpha+\varphi$. Therefore, the sum of the angles of triangle $A M C$ is $(2 \alpha+\varphi)+2 \varphi+\alpha=3(\alpha+$\varphi). Consequently, $\angle C A N=\alpha+\varphi=60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_808033c11c6fe49d1738g-07.jpg?height=669&width=446&top_left_y=1055&top_left_x=835)

Answer

$60^{\circ}$.