[ Two pairs of similar triangles ] [ Quadratic equations. Vieta's theorem ]

In the angles $B$ and $C$ of triangle $ABC$, two circles with radii 2 and 3 are inscribed, touching the angle bisector of angle $A$ of the triangle.

Find this bisector, if the distance between the points where the circles touch $BC$ is 7.
Let a circle of radius 2 with center $O_{1}$ touch the bisector $A D$ and side $B C$ of triangle $A B C$ at points $M$ and $K$ respectively, and a circle of radius 3 with center $O_{2}$ touch $A D$ and $B C$ at points $N$ and $L$ respectively. Denote $\angle O_{1} D K=\alpha$. Since $D O_{1}$ and $D O_{2}$ are bisectors of adjacent angles, $\angle O_{1} D O_{2}=90^{\circ}, \angle O_{2} D L=90^{\circ}-$ $\angle O_{1} D K=90^{\circ}-\alpha, \angle D O_{2} L=\alpha$.

$A O_{1}$ and $A O_{2}$ are bisectors of equal angles $C A D$ and $B A D$, so $\angle O_{1} A M=\angle O_{2} A N$. Therefore, the right triangle $O_{1} A M$ is similar to the right triangle $O_{2} A N$, with the similarity ratio equal to $O_{1} M / O_{2} N=2 / 3$. In particular, $A M<A N$, so $D L=D N<D M=D K$.

Triangles $O_{1} D K$ and $L D O_{2}$ are also similar, from which $K D \cdot D L=O_{1} K \cdot O_{2} L=6$. Since $K D+D L=K L=7$, by Vieta's theorem $D M=K D=6$,

$D N=D L=1, M N=D M-D N=5$.

Since $A N=3 / 2 A M$, then $A M=2 M N=10$, and $A D=A M+M D=16$.

## Answer

16.