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 $4 \sqrt{2}$, $\sqrt{2}$, $\frac{8}{\sqrt{17}}$, and $\frac{8}{\sqrt{17}}$ respectively.
a) Find the ratio $K T: T M$.

b) Find the length of diagonal $L N$, given that $K M=10$.
Answer: $K T: T M=4, B D=\frac{50}{\sqrt{34}}$.

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 $k_{1}=\frac{\sqrt{17}}{\sqrt{2}}$ for the first pair and $k_{2}=\frac{4 \sqrt{2}}{\sqrt{17}}$ for the second pair.

Let $K T=\sqrt{17} x$. Then $N T=\frac{K T}{k_{1}}=x \sqrt{2} ; M T=\frac{N T}{k_{2}}=\frac{x \sqrt{17}}{4}, L T=\frac{K T}{k_{2}}=\frac{17 x}{4 \sqrt{2}}$.

Thus, $K T: T M=x \sqrt{17}: \frac{x \sqrt{17}}{4}=4: 1$.

If $K M=10$, then $x \sqrt{17}+\frac{x \sqrt{17}}{4}=10, x=\frac{8}{\sqrt{17}}$. Therefore, $L N=x \sqrt{2}+\frac{17 x}{4 \sqrt{2}}=\frac{25 x}{4 \sqrt{2}}=\frac{25 \sqrt{2}}{\sqrt{17}}$.

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