3. Given an isosceles triangle $ABC$ where $\overline{AC}=\overline{BC}$ and the height dropped from vertex $C$ is equal to half of the angle bisector of the angle at vertex $A$. Find the angle at vertex $C$.
Solution. Let $C C_{1}$ be the height dropped from vertex $C$ and let $A A_{1}$ be the angle bisector of angle $\alpha$ at vertex $A$ (see diagram). Since triangle $A B C$ is isosceles, point $C_{1}$ is the midpoint of side $A B$ and $C C_{1}$ is the angle bisector of angle $\gamma$ at vertex $C$. Let $C_{2}$ be a point on line $C C_{1}$ such that $\overline{C C_{1}}=\overline{C_{1} C_{2}}$ (see diagram). From the fact that the diagonals of quadrilateral $A C_{2} B C$ bisect each other, it follows that this quadrilateral is a parallelogram. Since $A C_{2} \| C A_{1}$ and $\overline{A A_{1}}=\overline{C C_{2}}$, quadrilateral $A C_{2} A_{1} C$ is an isosceles trapezoid, from which it follows that $\frac{\gamma}{2}=\measuredangle C_{2} C A_{1}=\measuredangle C A_{1} A$. From triangles $A C_{1} C$ and $A A_{1} C$, we get $\gamma=90^{\circ}-\frac{\gamma}{2}$ and $\frac{\gamma}{2}=180^{\circ}-\gamma-\frac{\gamma}{2}$ respectively. Then from the equation

$$
90^{\circ}-\frac{\gamma}{2}=2\left(180^{\circ}-\gamma-\frac{\gamma}{2}\right)
$$

we obtain that $\gamma=108^{\circ}$.

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