The bisector of the angle [

In triangle $ABC$, all sides of which are different, the bisector of the external angle adjacent to angle $ACB$ intersects the extension of side $BA$ at point $D$ ($A$ is between $B$ and $D$). It is known that $BD - BC = m$, $AC + AD = n$. Find $CD$.

#
Place segment $B C_{1}$, equal to $B C$, on ray $B A$, and segment $A C_{2}$, equal to $A C$, on ray $A B$. Triangles $D C_{1} C$ and $D C_{2} C$ are similar.

## Solution

Place segment $B C_{1}$, equal to $B C$, on ray $B A$, and segment $A C_{2}$, equal to $A C$, on ray $A B$. Then $D C_{1}=m, D C_{2}=n$, and $C C_{1} B$ and $C C_{2} A$ are isosceles triangles.

Let $\alpha, \beta$, and $\gamma$ be the angles of triangle $A B C$. Then $\angle D C_{1} C=180^{\circ}-\left(90^{\circ}-\beta / 2\right)=90^{\circ}+\beta / 2, \angle D C C_{2}=\angle D C A + \angle A C C_{2}=\left(90^{\circ}-\gamma / 2\right)+1 / 2\left(180^{\circ}-\alpha\right)=90^{\circ}+\beta / 2$. Therefore, triangles $D C_{1} C$ and $D C C_{2}$ are similar by two angles. Hence, $C D: C_{2} D=C_{1} D: C D$, from which $C D^{2}=C_{1} D \cdot C_{2} D=m n$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6ed8f1205731b80a0039g-45.jpg?height=351&width=912&top_left_y=1511&top_left_x=573)

Answer

$\sqrt{m n}$.