4. A right triangle $ABC$ with hypotenuse $AB$ is inscribed in a circle. The feet of the perpendiculars from points $A, B$ to the tangent $k$ of this circle at point $C$ are denoted as $D, E$. Express the length of segment $DE$ in terms of the lengths of the legs of triangle $ABC$.


SOLUTION. Let the legs of the right triangle $ABC$ be denoted by $a, b$ and the opposite angles by $\alpha, \beta$. The midpoint of the hypotenuse $AB$ (the center of the circumscribed circle) is denoted by $O$ (Fig. 1).

The height $v = CP$ divides the triangle $ABC$ into triangles $ACP$ and $CBP$, which are similar to triangle $ABC$ according to the AA similarity criterion $\left(\alpha + \beta = 90^{\circ}\right)$. The segment $OC$ is perpendicular to $DE$ and $|OC| = |OA| = r$ (the radius of the circumscribed circle). Therefore, $|\angle OCA| = |\angle OAC| = \alpha$ and $|\angle DCA| = 90^{\circ} - |\angle OCA| = \beta$.

The right triangles $ACP$ and $ACD$ with the common hypotenuse $AC$ are thus congruent in the angles at vertex $C$. Therefore, they are congruent, and even symmetric with respect to the line $AC$. Similarly, triangles $CBP$ and $CBE$ are symmetric with respect to $BC$. Thus, $|CD| = |CE| = v$, and therefore $|DE| = 2v = 2ab / \sqrt{a^2 + b^2}$, since from the double expression for the area of triangle $ABC$ it follows that $v = ab / |AB|$, where $|AB| = \sqrt{a^2 + b^2}$.

Note. Instead of the double expression for the area, we can use the similarity of triangles $CBP$ and $ABC: \sin \alpha = |CP| / |AC| = |BC| / |AB|$.

![](https://cdn.mathpix.com/cropped/2024_04_17_5c1ce8f446d3f7622ca5g-2.jpg?height=483&width=665&top_left_y=1689&top_left_x=387)

Fig. 1

![](https://cdn.mathpix.com/cropped/2024_04_17_5c1ce8f446d3f7622ca5g-2.jpg?height=557&width=666&top_left_y=1569&top_left_x=1092)

Fig. 2