Problem 5. Determine the angles in the isosceles triangle $ABC$, in which the height dropped to the base is half the length of the angle bisector of one of the angles at the base.
Solution. Let $\triangle A B C$ have sides $\overline{A C}=\overline{B C}, \overline{C C_{1}}$ be the altitude dropped to the base $A B$, and $\overline{A A_{1}}$ be the length of the angle bisector of $\measuredangle B A C=\alpha$. According to the condition

![](https://cdn.mathpix.com/cropped/2024_06_05_a1dcf8241dfa57377282g-07.jpg?height=281&width=603&top_left_y=595&top_left_x=876)

$$
\overline{A A_{1}}=2 \overline{C C_{1}} \text { and } \measuredangle B A A_{1}=\frac{\alpha}{2} .
$$

Let $D$ be the midpoint of $B A_{1}$. Then $C_{1} D \| A A_{1}$ and $\overline{C_{1} D}=\frac{\overline{A A_{1}}}{2}=\overline{C C_{1}}$, so $\measuredangle B C_{1} D=\measuredangle B A A_{1}=\frac{\alpha}{2}$. Therefore, $\triangle C C_{1} D$ is isosceles and

$$
\measuredangle D C C_{1}=\measuredangle C_{1} D C=90^{\circ}-\alpha
$$

so it must be true that

$$
\measuredangle D C_{1} C=180^{\circ}-2\left(90^{\circ}-\alpha\right)=2 \alpha
$$

Finally,

$$
\measuredangle D C_{1} C+\measuredangle B C_{1} D=2 \alpha+\frac{\alpha}{2}=90^{\circ}
$$

from which it follows that $\alpha=36^{\circ}$ and $\beta=180^{\circ}-2 \alpha=108^{\circ}$.

## VIII Department