In the corners $B$ 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 being $O_{1} M / O_{2} N=2 / 3$. 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 $A D = A M + M D = 16$.