[ $\underline{\text { The ratio in which the bisector divides the side. }}$

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 ratio $\frac{A B}{A C}$, if $\frac{B D}{B E}=\frac{3}{5}$. # ]
By the property of the bisectors of the internal and external angles of a triangle, $\frac{A B}{A C}=\frac{B D}{C D}=\frac{B E}{C E}$. Let point $E$ lie on the extension of side $B C$ beyond point $B$. Suppose $B D=3 x, B E=5 x, D C=y$. Then

$$
\frac{3 x}{y}=\frac{A B}{A C}=\frac{5 x}{B x+y},
$$

from which we find that $y=12 x$. Therefore,

$$
\frac{A B}{A C}=\frac{3 x}{y}=\frac{3 x}{12 x}=\frac{1}{4}
$$

If point $E$ lies on the extension of side $B C$ beyond point $C$, then similarly we find that $\frac{A B}{A C}=\frac{\mathbf{4}}{\mathbf{1}}$.

## Answer

$1: 4$ or $4: 1$.