7. A quadrilateral $K L M N$ is inscribed in a circle with diagonals $K M$ and $L N$, which intersect at point $T$. The bases of the perpendiculars dropped from point $T$ to the sides of the quadrilateral lie on these sides. The distances from point $T$ to the sides $K L, L M, M N, N K$ are $5 \sqrt{2}$, $\sqrt{2}$, $5 \sqrt{\frac{2}{13}}$, and $5 \sqrt{\frac{2}{13}}$ respectively.
a) Find the ratio $K T: T M$.

b) Find the length of diagonal $L N$, given that $K M=12$.
Answer: $K T: T M=5, L N=\frac{36}{\sqrt{13}}$.

Solution. Since inscribed angles subtended by the same arc are equal, $\angle T L K=$ $\angle T M N, \angle T K L=\angle T N M$. Therefore, triangles $T K L$ and $T N M$ are similar. Similarly, it can be proven that $\triangle T K N \sim \triangle T L M$. Corresponding elements of similar figures are proportional to the similarity coefficient. In this case, the corresponding elements are the heights drawn from vertex $T$. From this, we find that the similarity coefficient is $k_{1}=$ $\sqrt{13}$ for the first pair and $k_{2}=\frac{5}{\sqrt{13}}$ for the second pair.

Let $K T=\sqrt{13} x$. Then $N T=\frac{K T}{k_{1}}=x ; M T=\frac{N T}{k_{2}}=\frac{x \sqrt{13}}{5}, L T=\frac{K T}{k_{2}}=\frac{13 x}{5}$.

Thus, $K T: T M=x \sqrt{13}: \frac{x \sqrt{13}}{5}=5: 1$.

If $K M=12$, then $x \sqrt{13}+\frac{x \sqrt{13}}{5}=12, x=\frac{10}{\sqrt{13}}$. Therefore, $L N=x+\frac{13 x}{5}=\frac{18 x}{5}=\frac{36}{\sqrt{13}}$.