[ Angles subtended by equal arcs and equal chords]

[ Auxiliary similar triangles $\quad]$

From the vertex of the obtuse angle $A$ of triangle $A B C$, a height $A D$ is dropped. From point $D$, a circle with radius equal to $A D$ is described, intersecting the sides of triangle $A B$ and $A C$ at points $M$ and $N$ respectively. Find the side $A C$, if it is known that $A B=c, A M=m$ and $A N=n$.
Prove that triangles $A M N$ and $A C B$ are similar.

## Solution

Extend $A D$ to intersect the given circle at point $P$. Then $\angle A M N = \angle A P N$, and since $\angle A N P = 90^{\circ}$, it follows that $\angle A P N = \angle C$. Therefore, triangles $A M N$ and $A C B$ are similar. Consequently, $A M : A C = A N : A B$. Hence, $A C = A B^{A M} / A N = m c / n$.

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-31.jpg?height=426&width=872&top_left_y=1348&top_left_x=593)

## Answer

$\mathrm{mc} / \mathrm{n}$.