10.235. In an isosceles triangle, the angle at the base is $72^{\circ}$, and the bisector of this angle has a length of $m$. Find the lengths of the sides of the triangle.
Solution.

Let in $\triangle ABC$, $AB=BC$, $\angle BAC=\angle BCA=72^{\circ}$, $AD$ is the bisector of $\angle BAC$, $AD=m$ (Fig. 10.43). $\angle BAD=\angle DAC=36^{\circ}$, $\angle ABC=180^{\circ}-2\angle BAC=36^{\circ}$, $\angle ADC=180^{\circ}-(\angle DAC+\angle BCA)=72^{\circ}$. Therefore, $\triangle ADB$ and $\triangle CAD$ are isosceles, $BD=AD=AC=m$. Let $AB=BC=x$. Then $DC=x-m$. Since $AD$ is the bisector of $\triangle ABC$, $\frac{BD}{DC}=\frac{AB}{AC}$; $\frac{m}{x-m}=\frac{x}{m}$; $x^2 - xm - m^2 = 0$, and since $x>0$, $x=\frac{m(1+\sqrt{5})}{2}$.

Answer: $m ; \frac{m(1+\sqrt{5})}{2}$.