In a triangle $ABC, AM$ is a median, $BK$ is a bisectrix, $L=AM\cap BK$. It is known that $BC=a, AB=c, a>c$.
Given that the circumcenter of triangle $ABC$ lies on the line $CL$, find $AC$

I. Voronovich

1. Given that \( AM \) is a median, \( BK \) is an angle bisector, and \( L = AM \cap BK \), we need to find \( AC \) given that the circumcenter of triangle \( ABC \) lies on the line \( CL \).

2. We start by using the area ratio property:
   \[
   \frac{S_{\triangle ACL}}{S_{\triangle LCM}} = \frac{AC \sin \angle ACL}{CM \sin \angle LCM} = \frac{AL}{LM} = \frac{AB}{BM}
   \]
   This implies:
   \[
   AC \sin \angle ACL = AB \sin \angle LCM
   \]

3. Since the circumcenter of triangle \( ABC \) lies on the line \( CL \), we have:
   \[
   \angle ACL = 90^\circ - \angle ABC \quad \text{and} \quad \angle LCM = 90^\circ - \angle CAB
   \]

4. Substituting these angles into the equation, we get:
   \[
   AC \sin (90^\circ - \angle ABC) = AB \sin (90^\circ - \angle CAB)
   \]
   Using the trigonometric identity \( \sin(90^\circ - x) = \cos(x) \), we have:
   \[
   AC \cos \angle ABC = AB \cos \angle CAB
   \]

5. Let \( AB = c \), \( BC = a \), and \( AC = b \). Then:
   \[
   b \cos \angle ABC = c \cos \angle CAB
   \]

6. Using the Law of Cosines in \( \triangle ABC \):
   \[
   \cos \angle ABC = \frac{a^2 + c^2 - b^2}{2ac} \quad \text{and} \quad \cos \angle CAB = \frac{b^2 + c^2 - a^2}{2bc}
   \]

7. Substituting these into the equation, we get:
   \[
   b \cdot \frac{a^2 + c^2 - b^2}{2ac} = c \cdot \frac{b^2 + c^2 - a^2}{2bc}
   \]

8. Simplifying, we have:
   \[
   b \cdot \frac{a^2 + c^2 - b^2}{2ac} = \frac{c(b^2 + c^2 - a^2)}{2bc}
   \]
   \[
   b(a^2 + c^2 - b^2) = c(b^2 + c^2 - a^2)
   \]

9. Multiplying both sides by \( 2ac \), we get:
   \[
   b^2(a^2 + c^2 - b^2) = ac(b^2 + c^2 - a^2)
   \]

10. Rearranging terms, we have:
    \[
    b^4 - b^2(a^2 + c^2 - ac) - ac(a^2 - c^2) = 0
    \]

11. This is a quadratic equation in \( b^2 \). Solving for \( b^2 \), we can find \( b \).

The final answer is \( \boxed{ AC = b } \).