## Task B-3.5.

In an isosceles triangle $ABC$ with an obtuse angle at vertex $C$, the foot of the altitude on side $\overline{BC}$ is point $D$. Determine the angles of triangle $ABC$, if $\frac{|AB|+|BD|}{|AC|+|CD|}=\frac{2 \sqrt{3}+3}{3}$.
## First Solution.

Let $a=|A B|, b=|A C|=|B C|$, and $|C D|=x$.

![](https://cdn.mathpix.com/cropped/2024_05_30_d3836afd55956f7fff9cg-16.jpg?height=500&width=885&top_left_y=395&top_left_x=494)

With these notations, the given equality becomes $\frac{a+b+x}{b+x}=\frac{2 \sqrt{3}+3}{3}$. From this, we have

$$
\frac{a}{b+x}=\frac{a+b+x}{b+x}-1=\frac{2 \sqrt{3}+3}{3}-1=\frac{2 \sqrt{3}}{3}
$$

or $\frac{b+x}{a}=\frac{\sqrt{3}}{2}$.

Notice that $\cos \beta=\frac{b+x}{a}$,

so from the previous result, we get $\cos \beta=\frac{\sqrt{3}}{2}$, hence $\beta=30^{\circ}$.

Since $\alpha+2 \beta=180^{\circ}$, we obtain $\alpha=120^{\circ}$.

Therefore, the angles of triangle $A B C$ are $30^{\circ}, 30^{\circ}, 120^{\circ}$.