3. Given $\triangle A B C$ such that $\measuredangle C A B=3 \measuredangle A B C$. Let $L$ be the intersection point of the angle bisector of $\measuredangle A C B$ with side $A B$, and let $P_{\triangle A L C}: P_{\triangle L B C}=1: 2$,
where $P_{\triangle A L C}$ and $P_{\triangle L B C}$ are the areas of $\triangle A L C$ and $\triangle L B C$, respectively. Determine the measures of the angles of $\triangle A B C$
Solution. For $\triangle A B C$, we will use standard notations. Then $\alpha=\measuredangle C A B=3 \measuredangle A B C=3 \beta$. Let $M$ be a point on $B C$ such that $\measuredangle B A M=\beta$. Thus, $\triangle A B M$ is isosceles and $\overline{M A}=\overline{M B}$. Furthermore,

$$
\begin{aligned}
& \measuredangle M A C=\measuredangle B A C-\measuredangle B A M=2 \beta \text{ and } \\
& \measuredangle A M C=\measuredangle B A M+\measuredangle A B M=2 \beta
\end{aligned}
$$

which means $\triangle A M C$ is isosceles and $\overline{A C}=\overline{M C}$.

Let $S$ be the intersection of $A M$ and $C L$. Now $C S$ is the angle bisector of $\measuredangle A C B$ and $\triangle A M C$ is

![](https://cdn.mathpix.com/cropped/2024_06_05_b088eb3db01806ad63fdg-07.jpg?height=337&width=485&top_left_y=538&top_left_x=995)
isosceles with vertex $C$, so $C S$ is the altitude in this triangle, which means $C L$ is the perpendicular bisector of side $A M$ in $\triangle A M C$, and thus also in $\triangle A M L$. This means $\triangle A M L$ is isosceles, i.e., $\overline{A L}=\overline{M L}$.

Furthermore, $\overline{A C}=\overline{M C}, \overline{A L}=\overline{M L}$, and $\overline{C L}=\overline{C L}$, so $\triangle A L C \cong \triangle M L C$, i.e., $P_{\triangle A L C}=P_{\triangle M L C}$. From this equality and the condition of the problem, it follows that

$$
2 P_{\triangle A L C}=P_{\triangle L B C}=P_{\triangle L B M}+P_{\triangle L M C}=P_{\triangle L B M}+P_{\triangle A L C}
$$

i.e., $P_{\triangle A L C}=P_{\triangle L B M}$, so $P_{\triangle L B M}=P_{\triangle M L C}$. However, in triangles $L B C$ and $L M C$, the altitudes from vertex $L$ are equal, so from the last equality, it follows that the corresponding bases are equal, i.e., $\overline{M B}=\overline{M C}$. Thus, $\overline{A C}=\overline{M C}=\overline{M B}=\overline{M A}$, i.e., $\triangle A M C$ is equilateral. Therefore, $\measuredangle A C B=\measuredangle A C M=\measuredangle M A C=60^{\circ}$, and since $\measuredangle M A C=2 \beta$, we get $\beta=30^{\circ}$ and finally $\alpha=3 \beta=90^{\circ}$.