Similarity criteria [ Properties and characteristics of an isosceles triangle. ]

In an acute triangle $ABC$, a point $M$ is taken on the altitude $AD$, and a point $N$ is taken on the altitude $BP$ such that angles $BMC$ and $ANC$ are right angles. The distance between points $M$ and $N$ is $4 + 2\sqrt{3}$, and angle $MCN$ is $30^\circ$. Find the bisector $CL$ of triangle $CMN$.
Since $M D$ and $N P$ are the altitudes of the right triangles $B M C$ and $A N C$, drawn from the vertices of the right angles, then $C M^{2}=B C \cdot C D=A C \cdot C P=C N^{2}$.

## Solution

From the similarity of the right triangles $A C D$ and $B C P$, it follows that $C D: A C=C P: B C$, hence $B C \cdot C D=$ $C P \cdot A C$.

Since $M D$ and $N P$ are the altitudes of the right triangles $B M C$ and $A N C$, drawn from the vertices of the right angles, then $C M^{2}=B C \cdot C D=A C \cdot C P=C N^{2}$, therefore $C M=C N$.

The bisector $C L$ of the isosceles triangle $C M N$ is its altitude and median, thus,

$C L=M L \operatorname{ctg} \angle M C L=1 / 2 M N \operatorname{ctg} 15^{\circ}=\frac{(2+\sqrt{3})\left(1+\cos 30^{\circ}\right)}{\sin 30^{\circ}}=(2+\sqrt{3})^{2}=7+4 \sqrt{3}$.

## Answer

$7+4 \sqrt{3}$.