2. Find the ratio of the length of the base to the length of the leg in an isosceles triangle with an angle at the base of $72^{\circ}$.
Solution. Let the given triangle be $ABC, \overline{AC} = \overline{BC}$ and $\measuredangle ABC = \measuredangle BAC = 72^{\circ}$, as shown in the diagram on the right. We will denote $\overline{AC} = \overline{BC} = a$ and $\overline{AB} = c$. Let $AL (L \in BC)$ be the angle bisector at vertex $A$. Then the triangles $ABC$, $CAL$, and $BLA$ are similar (they have the same angles), so according to this, $\overline{AL} = \overline{AB} = \overline{CL} = c$. From the similarity of $ABC$ and $BLA$, we get

![](https://cdn.mathpix.com/cropped/2024_06_05_ff7f87db155b1f76013dg-4.jpg?height=406&width=358&top_left_y=234&top_left_x=1143)

$$
\frac{\overline{AB}}{\overline{AC}} = \frac{\overline{BL}}{\overline{AB}} \Leftrightarrow \frac{c}{a} = \frac{a - c}{c}
$$

which simplifies to $c^2 + ac - a^2 = 0$. We have $\left(\frac{c}{a}\right)^2 + \frac{c}{a} - 1 = 0$ and therefore $\frac{c}{a} = \frac{-1 \pm \sqrt{5}}{2}$. Thus, the desired ratio is $\frac{c}{a} = \frac{-1 + \sqrt{5}}{2}$.