[ The ratio in which the bisector divides the side. ]

From vertex $A$ of triangle $A B C$, the bisectors of the internal and external angles are drawn, intersecting line $B C$ at points $D$ and $E$ respectively. Find the radius of the circumcircle of triangle $A D E$, if $B C=a$ and $\frac{A B}{A C}=\frac{2}{3}$.
By the property of the bisectors of the internal and external angles of a triangle, $\frac{B D}{C D}=\frac{B E}{C E}=\frac{A B}{A C}=\frac{2}{3}$. Let point $E$ lie on the extension of side $B C$ beyond point $B$. Suppose $B D=2 t, D C=3 t, B E=y$. Then $5 t=B C=a, t=\frac{\frac{2}{5}}{5}$,

$$
\frac{y}{y+5 t}=\frac{A B}{A C}=\frac{2}{3}
$$

from which we find that $y=10 t$. Therefore,

$$
D E=B E+B D=y+2 t=10 t+2 t=12 t,
$$

and since the angle between the bisectors of adjacent angles is $90^{\circ}$, $D E$ is the hypotenuse of the right triangle $A D E$, so the radius of the circle circumscribed around this triangle is

$$
\frac{1}{2} D E=6 t=6 \cdot \frac{\frac{6}{5}}{5}=\frac{6}{5} a \text {. }
$$

Similarly for the case when point $E$ lies on the extension of side $B C$ beyond point $C$.

## Answer

$\frac{6}{5} a$