3. On the hypotenuse $A B$ of an isosceles right triangle $A B C$, points $K$ and $L$ are marked such that $A K: K L: L B=1: 2: \sqrt{3}$. Find $\angle K C L$.
Answer: $45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_66368caef28b3de8e403g-30.jpg?height=382&width=392&top_left_y=400&top_left_x=912)

Solution 1. Let $A K=1$. Then $K L=2$ and $L B=\sqrt{3}$, so $A B=3+\sqrt{3}$ and $A C=B C=\frac{3+\sqrt{3}}{\sqrt{2}}$. Therefore,

$$
A C \cdot B C=6+3 \sqrt{3}=3(2+\sqrt{3})=A L \cdot K B, \quad \text { or } \quad \frac{K B}{B C}=\frac{C A}{A L} .
$$

Since $\angle K B C=45^{\circ}=\angle C A L$, triangles $C B K$ and $L A C$ are similar. Therefore, $\angle B C K=\angle A L C$ and

$$
\angle A L C=\angle B C L+\angle L B C=\angle B C L+45^{\circ}=\angle B C K-\angle K C L+45^{\circ}=\angle A L C-\angle K C L+45^{\circ},
$$

from which $\angle K C L=45^{\circ}$.

Solution 2. Extend triangle $A B C$ to form a square $A B C D$, and let $M$ be the intersection point of lines $B D$ and $C L$ (see the figure). Let $a=A C$. Triangles $A L C$ and $B L M$ are similar with a ratio of $\sqrt{3}$. Therefore,

$$
B M=\frac{a}{\sqrt{3}}, \quad C M=\sqrt{a^{2}+\frac{a^{2}}{3}}=\frac{2 a}{\sqrt{3}}, \quad \frac{A B}{C M}=\sqrt{\frac{3}{2}}
$$

Moreover,

$$
B L=\frac{A B}{\sqrt{3}+1}, \quad C L=\frac{C M \cdot \sqrt{3}}{\sqrt{3}+1}, \quad \frac{C L}{B L}=\sqrt{3} \cdot \frac{C M}{A B}=\sqrt{2}
$$

and also

$$
K L=\frac{2 A B}{\sqrt{3}+3}, \quad M L=\frac{C M}{\sqrt{3}+1}, \quad \frac{K L}{M L}=\frac{2}{\sqrt{3}} \cdot \frac{A B}{C M}=\sqrt{2}
$$

Thus, $\frac{C L}{B L}=\frac{K L}{M L}$, and triangles $B L M$ and $C L K$ are similar. Therefore, $\angle K C L=\angle M B L=45^{\circ}$.