In an acute-angled triangle $A B C$, altitudes $A A_{1}$ and $B B_{1}$ are drawn. The bisector of the exterior angle at vertex $C$ intersects the lines $A B$ and $A_{1} B_{1}$ at points $L$ and $K$ respectively. It turns out that $C L=2 C K$. Find the angle $C$.
As is known, triangle $C A_{1} B_{1}$ is similar to triangle $C A B$ with a similarity coefficient of $\cos \angle C$.

Let's assume that point $L$ lies on the extension of side $A B$ beyond point $B$. We will prove that point $C$ lies between $K$ and $L$. Suppose this is not the case. Then, if $P$ is a point on the extension of side $A C$ beyond point $C$, and $Q$ is a point on the extension of side $A B$ beyond point $A$, then $\angle A_{1} C K = \angle L C P$,

$\angle C A_{1} K = 180^{\circ} - \angle C A_{1} B_{1} = 180^{\circ} - \angle C A B = \angle C A Q$, which means the sum of two angles of triangle $C A_{1} K$ equals the sum of two exterior angles of triangle $A C L$, i.e., more than $180^{\circ}$. Contradiction.

Notice that the angles of triangles $C A L$ and $C A_{1} K$ at vertex $C$ are equal. Moreover, $\angle A = \angle C A_{1} B_{1}$. Therefore, these triangles are similar. Hence,

$\cos \angle C = C A_{1 / C A} = C K_{/} / C L = 1 / 2$

## Answer

$60^{\circ}$.