On the sides $A B$ and $A C$ of the isosceles triangle $A B C$, points $N$ and $M$ are located respectively, such that $A N=N M=M B=B C$.

Find the angles of triangle $A B C$.

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Apply the exterior angle theorem of a triangle.

## Solution

Let $\angle A=\alpha$. From the isosceles triangle $A N M$, we find that $\angle A M N=\angle B A C=\alpha$. Since $B N M$ is the exterior angle of triangle $A N M$, then

$\angle B N M=2 \alpha$. Therefore, $\angle M B N=\angle B N M=2 \alpha$, and since $B M C$ is the exterior angle of triangle $A M B$, then $\angle B M C=$ $\angle B A C+\angle M B N=\alpha+2 \alpha=3 \alpha$. Thus,

$\angle B=\angle C=\angle B M C=3 \alpha$.

By the theorem of the sum of the angles of a triangle, $3 \alpha+3 \alpha+\alpha=180^{\circ}$, from which $\alpha=180^{\circ} / 7$.

## Answer

![](https://cdn.mathpix.com/cropped/2024_05_06_96d74c2697cd905bd948g-31.jpg?height=71&width=349&top_left_y=645&top_left_x=17)