[ Auxiliary similar triangles ] [ Angles subtending equal arcs and equal chords ]

In an acute-angled triangle $A B C$, side $A B$ is less than side $A C$, and $D$ is the point of intersection of the line $D B$, perpendicular to $A B$, and the line $D C$, perpendicular to $A C$. The line passing through point $B$ and perpendicular to $A D$ intersects $A C$ at point $M$. It is known that $A M=m, M C=n$. Find $A B$.
Triangles $A B M$ and $A C B$ are similar.

## Solution

Points $B$ and $C$ lie on the circle with diameter $A D$. Therefore, $\angle A B M=90^{\circ}-\angle M B D=\angle B D A=\angle C$.

Thus, triangles $A B M$ and $A C B$ are similar by two angles. Therefore, $A M: A B=A B: A C$. Hence, $A B^{2}=$ $A M \cdot A C=m(m+n)$

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-37.jpg?height=414&width=460&top_left_y=1488&top_left_x=799)

## Answer

$$
\sqrt{m(m+n)}
$$