Problem 4. In a right triangle $KLM$, a circle is constructed on the leg $KM$ as its diameter, which intersects the hypotenuse $KL$ at point $G$. A tangent to the circle is drawn through point $G$, intersecting the leg $ML$ at point $F$. Find the length of $FL$, if $KG=5$ and $LG=4$.
Answer: 3.

Solution. The solution is based on two simple observations. First, $\angle K G M=90^{\circ}$, since it subtends the diameter. Second, $F G$ and $F M$ are tangents to the circle from the condition, so $F G=F M$. Therefore, in the right triangle $M G L$, a point $F$ is marked on the hypotenuse $L M$ such that $F G=F M$; it is well known that then $F$ is the midpoint of this hypotenuse, $L F=L M / 2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4a99e6e32d5aecd6d775g-21.jpg?height=700&width=1317&top_left_y=621&top_left_x=358)

The solution can be completed in several ways; we will present two of them.

First method, height of a right triangle and the Pythagorean theorem. In the right triangle $K L M$, the height $M G$ divides the hypotenuse $K L$ into segments $K G=5$ and $L G=4$. We get that $M G^{2}=K G \cdot L G=5 \cdot 4$; by the Pythagorean theorem $L M^{2}=L G^{2}+M G^{2}=4^{2}+5 \cdot 4=36, L F=L M / 2=3$.

Second method, the tangent-secant theorem. According to the tangent-secant theorem for the tangent $L M$ and the secant $L K$, we have: $L M^{2}=L G \cdot L K=4 \cdot(4+5)=36, L F=L M / 2=3$.