# Answer: 8

## Solution:

Notice that the quadrilateral $L_{1} M_{2} L_{2} M_{1}$ is cyclic, since $\angle L_{1} M_{2} M_{1}=\angle L_{1} L_{2} M_{1}=$ $90^{\circ}$. Therefore, $\angle B M_{2} L_{2}=180^{\circ}-\angle L_{1} M_{2} L_{2}=\angle L_{2} M_{1} L_{1}=\angle B M_{1} L_{1}$. Similarly, $\angle B L_{2} M_{2}=\angle B L_{1} M_{1}$, so triangles $B L_{1} M_{1}$ and $B L_{2} M_{2}$ are similar, with point $L_{1}$ in the first triangle corresponding to point $L_{2}$ in the second. However, the problem states that they are similar in another way, which means these two triangles are isosceles. From this, we get $B L_{1}=B M_{1}=B L_{2}+L_{2} M_{1}$ and, by the Pythagorean theorem, we find $L_{1} L_{2}=\sqrt{B L_{1}^{2}-B L_{2}^{2}}=\sqrt{\left(B L_{2}+L_{2} M_{1}\right)^{2}-B L_{2}^{2}}$.
# Problem 7. (3 points)

From point $K$ on side $A C$ of triangle $A B C$, perpendiculars $K L_{1}$ and $K M_{1}$ were dropped to sides $A B$ and $B C$ respectively. From point $L_{1}$, a perpendicular $L_{1} L_{2}$ was dropped to $B C$, and from point $M_{1}$, a perpendicular $M_{1} M_{2}$ was dropped to $A B$.

It turned out that triangles $B L_{1} M_{1}$ and $B L_{2} M_{2}$ are similar (point $L_{1}$ in the first triangle corresponds to point $M_{2}$ in the second). In addition, $B L_{2}=6$ and $L_{2} M_{1}=4$. Find $L_{1} L_{2}$.