Problem 1041

Solution. Let $M D \cap A B=\{P\}$ and $M C \cap A B=\{Q\}$ (make a drawing). Now $\triangle P A M \sim \triangle D E M$ and $\triangle M K L \sim \triangle M A B$, so $\frac{\overline{M K}}{\overline{M A}}=\frac{\overline{D K}}{\overline{P A}}$ and $\frac{\overline{M K}}{\overline{M A}}=\frac{\overline{K L}}{\overline{A B}}$. From the last equalities, it follows that

$$\frac{\overline{A B}}{\overline{A P}}=\frac{\overline{K L}}{\overline{D K}}$$

On the other hand, for the right triangles $D A P$ and $Q C B$, $\measuredangle D P A=\measuredangle B C Q$ holds, as angles with perpendicular sides, so they are similar. According to this, $\frac{\overline{B Q}}{\overline{B C}}=\frac{\overline{A D}}{\overline{A P}}$, and since $A B C D$ is a square, we get $\frac{\overline{B Q}}{\overline{B C}}=\frac{\overline{A B}}{\overline{A P}}$. From the last equality and (1), it follows that

$$\frac{\overline{B Q}}{\overline{B C}}=\frac{\overline{K L}}{\overline{D K}}$$

On the other hand, triangles $M L C$ and $M B Q$ are similar, so $\frac{\overline{L C}}{\overline{B Q}}=\frac{\overline{M L}}{\overline{M B}}$ and since $\frac{\overline{M L}}{\overline{M B}}=\frac{\overline{K L}}{\overline{A B}}$, we get $\frac{\overline{L C}}{\overline{B Q}}=\frac{\overline{K L}}{\overline{A B}}=\frac{\overline{K L}}{\overline{B C}}$, i.e.

$$\frac{\overline{B Q}}{\overline{B C}}=\frac{\overline{L C}}{\overline{K L}}$$

Finally, from equalities (2) and (3), we get $\frac{\overline{K L}}{\overline{D K}}=\frac{\overline{L C}}{\overline{K L}}$, i.e. $\overline{K L}^{2}=\overline{D K} \cdot \overline{L C}$.