4. From the vertex of the obtuse angle $A$ of triangle $A B C$, the altitude $A D$ is dropped. A circle with center $D$ and radius $D A$ is drawn, intersecting sides $A B$ and $A C$ again at points $M$ and $N$ respectively. Find $A C$, if $A B=c, A M=m$ and $A N=n$.
Answer: $\frac{m c}{n}$.

Solution. We will prove that $A M \cdot A B = A N \cdot A C$. This can be done in different ways.

First method. In the right triangles $A D B$ and $A D C$, draw the altitudes $D P$ and $D Q$ respectively (see Fig. 10.4a). 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 get the required result.

Second method. We will prove that the quadrilateral $B M N C$ is cyclic, then the required equality will follow from the theorem of the segments of secants applied to point $A$ and the circumcircle of quadrilateral $B M N C$ (see Fig. 10.4b).

Let $\angle A N M = \alpha$, then $\angle A O M = 2 \alpha$ (inscribed and central angles subtending the same arc). Additionally, from the isosceles triangle $A D M: \angle M A D = 90^{\circ} - \alpha$, so $\angle A B C = \alpha$. From the equality $\angle A B C = \angle A N M$, it follows that $B M N C$ is cyclic.

After proving the required equality, it is sufficient to substitute the given data from the problem and obtain the answer.

Third method. Let the given circle intersect segments $B D$ and $C D$ at points $K$ and $L$ respectively, and let its radius be $R$ (see Fig. 10.4c). Then, by the theorem of the segments of secants: $B A \cdot B M = B L \cdot B K$, that is, $c(c - m) = B K(B K + 2 R)$. From triangle $A B D$ by the Pythagorean theorem: $c^{2} = (B K + R)^{2} + R^{2} = 2 R^{2} + B K^{2} + 2 B K \cdot R$. Therefore, $c(c - m) = c^{2} - 2 R^{2}$, from which $m = 2 R^{2}$.

By conducting a similar reasoning for side $A C$, we get $A C \cdot n = 2 R^{2}$. Then $A C = \frac{m c}{n}$.

Note that in this solution method, instead of the Pythagorean theorem, the cosine theorem can be applied to triangle $B A K$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ca89b09c7a9d835f2b71g-2.jpg?height=474&width=734&top_left_y=183&top_left_x=1186)

Grading criteria.
![](https://cdn.mathpix.com/cropped/2024_05_06_ca89b09c7a9d835f2b71g-2.jpg?height=1036&width=752&top_left_y=732&top_left_x=1188)

"+" - a complete and justified solution is provided

“士” - a generally correct reasoning is provided, containing minor gaps or inaccuracies (e.g., $m$ and $n$ are confused)

“耳” - the plan of the solution is correct and the correct answer is obtained, but some of the used facts are not proven (e.g., it is used but not proven that quadrilateral $BMNC$ is cyclic)

"干" - the plan of the solution is correct, but the solution itself contains errors or is not completed

“Ғ” - there is no clear plan of the solution, but some significant facts are justified, from which the solution can be obtained

“-" - only the answer is provided

“-" - the problem is not solved or solved incorrectly