10.236. In an isosceles triangle, the angle at the vertex is $36^{\circ}$, and the bisector of the angle at the base is $\sqrt{20}$. Find the lengths of the sides of the triangle.
Solution.

Refer to the diagram from the previous problem. In $\triangle ABC$, $AB = BC$, $\angle ABC = 36^\circ$, $AD$ is the angle bisector of $\angle BAC$, $AD = \sqrt{20}$. $\angle BAC = \angle BCA = 72^\circ$, $\angle BAD = \angle DAC = 36^\circ$, $\angle ADC = \angle BAD + \angle ABC = 72^\circ$. Therefore, $\triangle ADB$ and $\triangle CAD$ are isosceles, $BD = AD = AC = \sqrt{20} = 2\sqrt{5}$. Let $AB = BC = x$. Then $DC = x - 2\sqrt{5}$. Since $AD$ is the angle bisector of $\triangle ABC$, $\frac{BD}{DC} = \frac{AB}{AC}$, $\frac{2\sqrt{5}}{x - 2\sqrt{5}} = \frac{x}{2\sqrt{5}}$, $x^2 - 2x\sqrt{5} - 20 = 0$ and, since $x > 0$, $x = 5 + \sqrt{5}$.

Answer: $2\sqrt{5}; 5 + \sqrt{5}$.