4.- The angle $A$ of the isosceles triangle $\mathrm{ABC}$ measures $2 / 5$ of a right angle, with its angles B and C being equal. The bisector of its angle C intersects the opposite side at point D. Calculate the measures of the angles of triangle BCD. Express the measure $a$ of side BC in terms of the measure $b$ of side AC, without using trigonometric ratios in the expression.
## Solution

With the data from the statement we have:

![](https://cdn.mathpix.com/cropped/2024_04_26_2f2bf6346a1193d58cb1g-2.jpg?height=563&width=400&top_left_y=1906&top_left_x=1456)

in triangle $\mathrm{ABC} \angle \mathrm{BAC}=36^{\circ} ; \angle \mathrm{ABC}=\angle \mathrm{ACB}=72^{\circ}$

in triangle $\mathrm{CBD} \quad \angle \mathrm{BCD}=36^{\circ} ; \angle \mathrm{CDB}=\angle \mathrm{BDC}=72^{\circ}$
in triangle $\mathrm{ADC} \quad \angle \mathrm{DAC}=\angle \mathrm{ACD}=72^{\circ} ; \angle \mathrm{ADC}=108^{\circ}$

therefore $\triangle \mathrm{BCD}$ and $\triangle \mathrm{ADC}$ are isosceles and also $\triangle \mathrm{BCD}$ is similar to $\triangle \mathrm{ABC}$.

For the sides we have: $\mathrm{DC}=\mathrm{AD}=\mathrm{a} ; \mathrm{BD}=\mathrm{b} - \mathrm{a}$.

Expressing the proportionality derived from the similarity above:

$$
\frac{\mathrm{b}-\mathrm{a}}{\mathrm{a}}=\frac{\mathrm{a}}{\mathrm{b}} \Leftrightarrow \mathrm{a}^{2}=\mathrm{b}^{2}-\mathrm{ab} \Leftrightarrow \mathrm{a}^{2}+\mathrm{ab}-\mathrm{b}^{2}=0 \Leftrightarrow\left(\frac{\mathrm{a}}{\mathrm{b}}\right)^{2}+\frac{\mathrm{a}}{\mathrm{b}}-1=0
$$

and solving it yields $\frac{\mathrm{a}}{\mathrm{b}}=\frac{\sqrt{5}-1}{2} \Leftrightarrow \mathrm{a}=\frac{(\sqrt{5}-1) \mathrm{b}}{2}$, which means that a is the golden section of $\mathrm{b}$.