[ Auxiliary similar triangles ] [Inscribed angle, subtending the diameter ]

In triangle $A B C$, angle $C$ is obtuse, $D$ is the intersection point of line $D B$, perpendicular to $A B$, and line $D C$, perpendicular to $A C$. The altitude of triangle $A D C$, drawn from vertex $C$, intersects $A B$ at point $M$. It is known that $A M=a, M B=b$. Find $A C$.
## Solution

Notice that points $C$ and $B$ lie on the circle with diameter $A D$.

Draw the altitude $C N$ of triangle $A D C$ and extend it to intersect the circle at point $K$. Since $A D$ is the diameter perpendicular to the chord $C K$,

$C N = N K$ and $A C = A K$. Therefore, $\angle A C K = \angle A K C = \angle B$.

Triangles $C A M$ and $B A C$ are similar by two angles. Thus, $A C: A B = A M: A C$, which gives $A C^2 = A M \cdot A B = a(a+b)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6ed8f1205731b80a0039g-21.jpg?height=472&width=506&top_left_y=370&top_left_x=776)

## Answer

$\sqrt{a(a+b)}$.