Problem 4. Given an isosceles triangle $ABC$, where $\overline{AC}=\overline{BC}$. On the leg $AC$ are chosen points $M$ and $N$ such that $\measuredangle M B A=\measuredangle C B N$ and $\overline{M N}=\overline{B M}$, with point $M$ between points $A$ and $N$. Determine $\measuredangle N B A$.
Solution. Let $\measuredangle B A C=\measuredangle C B A=\alpha$, $\measuredangle M B A=\measuredangle C B N=x$ and $\quad \measuredangle N B M=y$. Then $\alpha=2 x+y$. Triangle BNM is isosceles, so

$$
\measuredangle N B M=\measuredangle M N B=y .
$$

Furthermore, the angles of triangle $A B N$ are $2 x+y, x+y$ and $y$, so

$$
2 x+y+x+y+y=180^{\circ}
$$

from which we get $x+y=60^{\circ}$.

Finally, $\Varangle N B A=x+y=60^{\circ}$.

## VIII section

![](https://cdn.mathpix.com/cropped/2024_06_05_3a3538b5fdc92afac337g-09.jpg?height=658&width=534&top_left_y=267&top_left_x=932)