Problem 4. In a right triangle $ABC$, a circle is constructed on the leg $AC$ as its diameter, which intersects the hypotenuse $AB$ at point $E$. A tangent to the circle is drawn through point $E$, which intersects the leg $CB$ at point $D$. Find the length of $DB$, if $AE=6$, and $BE=2$.
Answer: 2.

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

![](https://cdn.mathpix.com/cropped/2024_05_06_4a99e6e32d5aecd6d775g-03.jpg?height=694&width=1313&top_left_y=470&top_left_x=360)

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 $A B C$, the height $C E$ divides the hypotenuse $A B$ into segments $A E=6$ and $B E=2$. We get that $C E^{2}=A E \cdot B E=6 \cdot 2$; by the Pythagorean theorem $B C^{2}=B E^{2}+C E^{2}=2^{2}+2 \cdot 6=16, B D=B C / 2=2$.

Second method, the tangent-secant theorem. According to the tangent-secant theorem for the tangent $B C$ and the secant $B A$, we have: $B C^{2}=B E \cdot B A=2 \cdot(2+6)=16, B D=B C / 2=2$.