## Task A-2.4.

In triangle $A B C$, $|A B|=|A C|$, and the angle bisector of $\varangle A B C$ intersects side $\overline{A C}$ at point $D$ such that $|B C|=|B D|+|A D|$. Determine the angles of this triangle.
## Solution.

Let's choose point $E$ on side $\overline{B C}$ such that $|B E|=|B D|$ and $|C E|=|A D|$.

![](https://cdn.mathpix.com/cropped/2024_05_30_7166a15529367baa5955g-10.jpg?height=489&width=1017&top_left_y=612&top_left_x=514)

Since $B D$ is the angle bisector of $\angle C B A$, we have $|C D|:|A D|=|B C|:|A B|$.

Therefore, $\frac{|C D|}{|C E|}=\frac{|C D|}{|A D|}=\frac{|B C|}{|A B|}=\frac{|C B|}{|C A|}$.

Triangles $A B C$ and $E D C$ share the same angle at vertex $C$ and have the same ratios of corresponding sides at that vertex, so they are similar. Thus, $\angle C E D = \angle C A B$.

Let $\angle A B C = 2x$. Then $\angle C B D = \angle E B D = x$, $\angle D B A = x$, and $\angle A C B = 2x$.

Furthermore, $\angle C E D = \angle B A C = 180^\circ - 4x$.

Since triangle $B E D$ is isosceles, we have $\angle D E B = 90^\circ - \frac{x}{2}$.

Since $\angle D E B$ and $\angle C E D$ are supplementary angles, we have

\[
\left(90^\circ - \frac{x}{2}\right) + \left(180^\circ - 4x\right) = 180^\circ
\]

from which it follows that $x = 20^\circ$.

Finally, the angles of the observed triangle are $\angle A B C = \angle A C B = 40^\circ$ and $\angle B A C = 100^\circ$.