Sharygin I.F.

Inside triangle $ABC$ with acute angles at vertices $A$ and $C$, a point $K$ is taken such that $\angle AKB = 90^{\circ}$ and $\angle CKB = 180^{\circ} - \angle C$.

In what ratio does the line $BK$ divide the side $AC$, if the height dropped to $AC$ divides this side in the ratio $\lambda$, counting from vertex $A$?
Continue the segment $B K$ until it intersects side $A C$ at point $M$. Then $\angle M K C=180^{\circ}-\angle C K B=\angle C=$ $\angle M C B$, so the isosceles triangles $M K C$ and $M C B$ are similar.

Let $B D$ be the height of triangle $A B C, D C=a, A D=\lambda a, D M=x$. Then $C M: K M=B M: C M$, or $K M \cdot B M=$ $C M^{2}=(a-x)^{2}$

Points $D$ and $K$ lie on the circle with diameter $A B$. Therefore, $K M \cdot B M=A M \cdot D M=(\lambda a+x) x$.

From the equation $(a-x)^{2}=(\lambda a+x) x$, we find that $x=\frac{a}{\lambda+2}$, from which $A M=\lambda a+x=\frac{a(\lambda+1)^{2}}{\lambda+2}, M C=a-x=$ $\frac{a(\lambda+1)}{\lambda+2}$.

Therefore, ${ }^{A M} / M C=\lambda+1$.

## Answer

$A M / M C=\lambda+1$