4. A perpendicular bisector is drawn to the lateral side $AC$ of an isosceles triangle $ABC$. It intersects the lateral side $AB$ at point $L$ and the extension of the base at point $K$. It turns out that the areas of triangles $ALC$ and $KBL$ are equal. Find the angles of the triangle.
Answer: $36^{\circ}, 72^{\circ}, 72^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_bd69057b1c245517edf6g-23.jpg?height=394&width=586&top_left_y=494&top_left_x=815)

Solution 1. Let $M$ and $N$ be the midpoints of segments $A C$ and $B C$ respectively, $\alpha=\angle A B C$. Then $M N$ is the midline of triangle $A B C$ and, therefore, $M N \| A B$. Since point $K$ lies on the perpendicular bisector of $A C$, segments $K A$ and $K C$ are equal, hence $\angle K A C=\angle K C A=\alpha$. By the condition,

$$
2=\frac{S_{K B L}}{S_{A M L}}=\frac{K L \cdot L B}{L M \cdot A L} \Longleftrightarrow \frac{L B}{A L}=2 \cdot \frac{L M}{K L}
$$

Since $K L$ is the angle bisector of triangle $A K B$, we have $\frac{L B}{A L}=\frac{K B}{A K}$. Moreover,

$$
2 \cdot \frac{L M}{K L}=2 \cdot \frac{B N}{K B}=\frac{B C}{K B}
$$

Then from these two relations,

$$
\frac{K B}{C K}=\frac{K B}{A K}=\frac{L B}{A L}=2 \cdot \frac{L M}{K L}=\frac{B C}{K B} \Longrightarrow K B^{2}=B C \cdot C K
$$

Triangles $A B C$ and $K A C$ are similar by two angles, hence

$$
\frac{C K}{A B}=\frac{A B}{B C} \Longrightarrow A B^{2}=B C \cdot C K
$$

Therefore, $K B=A B$, so triangle $A B K$ is isosceles. Note that

$$
\angle B A K=\angle C A K-\angle C A B=\alpha-\left(180^{\circ}-2 \alpha\right)=3 \alpha-180^{\circ}, \quad \angle A K B=180^{\circ}-2 \alpha
$$

We get $3 \alpha-180^{\circ}=180^{\circ}-2 \alpha$, from which $\alpha=72^{\circ}$. Thus, angles $A B C$ and $A C B$ are $72^{\circ}$, and angle $B A C$ is $36^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_bd69057b1c245517edf6g-23.jpg?height=385&width=582&top_left_y=2072&top_left_x=817)

Solution 2. Let $M$ be the midpoint of $A C$. Note that right triangles $A M L$ and $C M L$, as well as $A M K$ and $C M K$, are equal by two legs. Then $\angle A K M=\angle C K M$; denote the common value of these angles by $\alpha$. Triangles $B A C$ and $A K C$ are isosceles, so

$$
\angle A B C=\angle A C K=\angle K A C=90^{\circ}-\alpha \quad \text { and } \quad \angle A B K=90^{\circ}+\alpha .
$$

Triangles $A L K$ and $C L K$ are equal by two sides and an angle, hence

$$
\angle L C B=\angle L A K=180^{\circ}-\angle A B K-\angle A K B=90^{\circ}-3 \alpha \quad \text { and } \quad \angle L C M=\angle L A M=\angle K A C-\angle L A K=2 \alpha
$$

By the area formula for a triangle,

$$
\frac{1}{2} \cdot L K \cdot L B \cdot \sin \angle K L B=S_{K B L}=S_{A L C}=2 S_{A L M}=A L \cdot M L \cdot \sin \angle A L M=A L \cdot M L \cdot \sin \angle K L B
$$

Thus,

$$
2 \cdot \frac{A L}{L K}=\frac{L B}{M L}=\frac{L B}{L C \sin \angle L C M}=\frac{L B}{L C \sin 2 \alpha}
$$

By the Law of Sines for triangles $A L K$ and $B C L$,

$$
\frac{A L}{L K}=\frac{\sin \alpha}{\sin \left(90^{\circ}-3 \alpha\right)}=\frac{\sin \alpha}{\cos 3 \alpha} \quad \text { and } \quad \frac{L B}{L C}=\frac{\sin \left(90^{\circ}-3 \alpha\right)}{\sin \left(90^{\circ}-\alpha\right)}=\frac{\cos 3 \alpha}{\cos \alpha} \text {. }
$$

Substituting these relations into the previous equality, we get

$$
2 \cdot \frac{\sin \alpha}{\cos 3 \alpha}=2 \cdot \frac{A L}{L K}=\frac{L B}{L C \sin 2 \alpha}=\frac{\cos 3 \alpha}{\sin 2 \alpha \cdot \cos \alpha}
$$

Therefore,

$$
\cos ^{2} 3 \alpha=2 \sin \alpha \cos \alpha \sin 2 \alpha=\sin ^{2} 2 \alpha
$$

from which

$$
\cos 3 \alpha=\sin 2 \alpha \Longleftrightarrow 3 \alpha=90^{\circ}-2 \alpha \Longleftrightarrow \alpha=18^{\circ}
$$