[ Median to the hypotenuse ] [ Pythagorean Theorem (direct and inverse) ]

Outside the right triangle $A B C$ on its legs $A C$ and $B C$, squares $A C D E$ and $B C F G$ are constructed.

The extension of the altitude $C H$ of triangle $A B C$ intersects the line $D F$ at point $K$. Find $H K$, if the legs are equal to 2 and 3.
Prove that $C K$ is the median of the right triangle $C D F$.

## Solution

Let $A C=3, B C=2$. Denote $\angle B A C=\alpha$. Right triangles $A B C$ and $D F C$ are equal by two legs, so $D F=A B=\sqrt{13}$,

$\angle C D F=\angle A=\alpha, \angle D C K=\angle B C H=\alpha$, which means triangle $D K C$ is isosceles, $C K=D K$. Similarly, $C K = F K$. Therefore, $C K=\frac{1}{2} D F=\frac{\sqrt{13}}{2}$. Since $A C \cdot B C=A B \cdot C H$, then $C H=A C \cdot \frac{B C}{A B}=\frac{6}{\sqrt{13}}$.

Thus, $H K=C H+C K=\frac{6}{\sqrt{13}}+\frac{\sqrt{13}}{2}=\frac{25}{2 \sqrt{13}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_633abcb7b0e2bd041761g-14.jpg?height=423&width=638&top_left_y=2338&top_left_x=710)

## Answer

## [ Auxiliary similar triangles ] Problem 54854 Topics: [ Inscribed angle subtending a diameter $]$  [ Angles subtending equal arcs and equal chords]  Difficulty: $3+$  Grades: 8,9

In a circle, a diameter $M N$ and a chord $A B$ parallel to the diameter $M N$ are drawn. The tangent to the circle at point $M$ intersects the lines $N A$ and $N B$ at points $P$ and $Q$ respectively. It is known that $M P=p, M Q=q$. Find $M N$.

## Hint

Right triangles $M Q N$ and $M N P$ are similar.

## Solution

Arcs enclosed between parallel chords are equal, so the inscribed angles subtending these arcs are equal. Additionally, $\angle M B N=90^{\circ}$, so the angles $B M N$ and $M Q N$ are equal (each of them forms $90^{\circ}$ with the angle $M N Q$). Therefore, $\angle P N M=\angle A N M=\angle B M N=\angle M Q N$.

Thus, right triangles $M Q N$ and $M N P$ are similar by two angles. Therefore, $P M: M N=M N: Q M$, from which $M N^{2}=P M \cdot Q M=p q$.

## Answer

$\sqrt{p q}$