## [ Properties and characteristics of an isosceles triangle. ]

 [ The inscribed angle is half of the central angle ] [ The product of the lengths of the segments of the chords and the lengths of the segments of the secants ] [ Inscribed quadrilaterals (other) ]
[ Auxiliary similar triangles ]

From the vertex of the obtuse angle $A$ of triangle $A B C$, a height $A D$ is dropped. A circle with center $D$ and radius $D A$ is drawn, which intersects sides $A B$ and $A C$ again at points $M$ and $N$ respectively. Find $A C$, if $A B=c, A M=m$ and $A N=n$.
Let's prove that $A M \cdot A B = A N \cdot A C$ (i.e., $m c = n A C$).

First method. In the right triangles $A D B$ and $A D C$, draw the altitudes $D P$ and $D Q$ respectively (see the left figure). Then $A P \cdot A B = A D^2 = A Q \cdot A C$. Since triangles $A D M$ and $A D N$ are isosceles, we have $A P = \frac{1}{2} A M$ and $A Q = \frac{1}{2} A N$.

Substituting $A P$ and $A Q$ in the equation $A P \cdot A B = A Q \cdot A C$, we obtain the required result.
![](https://cdn.mathpix.com/cropped/2024_05_06_6409e68af5a1ba5e989dg-08.jpg?height=312&width=1104&top_left_y=2514&top_left_x=480)

Second method. Let $\angle A N M = \alpha$, then $\angle A O M = 2 \alpha$. From the isosceles triangle $A D M$, we get $\angle M A D = 90^\circ - \alpha$, so $\angle B = \alpha$. Therefore, the quadrilateral $B M N C$ is cyclic. The required equality follows from the theorem about the segments of secants drawn from point $A$ to its circumscribed circle (see the right figure).

Third method. Draw the diameter $A E$. From the similarity of the right triangles $A B D$ and $A E M$, it follows that $A B : A E = A D : A M$, that is,

$A M \cdot A B = A D \cdot A E$. Similarly, $A N \cdot A C = A D \cdot A E$.

## Answer

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