8,9 In an isosceles triangle $ABC (AB = BC)$, the bisectors $BD$ and $AF$ intersect at point $O$. The ratio of the area of triangle $DOA$ to the area of triangle $BOF$ is $\frac{3}{8}$. Find the ratio $\frac{AC}{AB}$.
Using the property of the bisector of a triangle, express the areas of the indicated triangles in terms of the area of triangle $ABC$.

## Solution

First method.

Let $A D=D C=a, A B=B C=b$. Let $S$ be the area of triangle $ABC$. By the property of the bisector of a triangle,

$$
\frac{B O}{O D}=\frac{A B}{A D}=\frac{b}{a}, \frac{B F}{F C}=\frac{A B}{A C}=\frac{b}{2 a}
$$

Therefore,

$$
\frac{O D}{B D}=\frac{a}{a+b}, \frac{B O}{B D}=\frac{b}{a+b}, \frac{B F}{B C}=\frac{b}{2 a+b}
$$

Then,

$$
\begin{gathered}
S_{\triangle \mathrm{AOD}}=\frac{O D}{B D} \cdot S_{\triangle \mathrm{ABD}}=\frac{a}{a+b} \cdot \frac{1}{2} S \\
S_{\triangle \mathrm{BOF}}=\frac{B O}{B D} \cdot \frac{B F}{B C} S_{\triangle \mathrm{DBC}}=\frac{b}{a+b} \cdot \frac{b}{2 a+b} \cdot \frac{1}{2} S
\end{gathered}
$$

From the problem statement, it follows that

$$
\frac{a}{a+b} \cdot \frac{1}{2} S=\frac{3}{8} \cdot \frac{b}{a+b} \cdot \frac{b}{2 a+b} \cdot \frac{1}{2} S
$$

or

$$
8 a(2 a+b)=3 b^{2}, 16 a^{2}+8 a b-3 b^{2}=0,16\left(\frac{a}{b}\right)^{2}+8\left(\frac{a}{b}\right)-3=0
$$

From this, we find that $\frac{a}{b}=\frac{1}{4}$. Therefore,

$$
\frac{A C}{A B}=\frac{2 a}{b}=\frac{1}{2}
$$

Second method.

Drop a perpendicular $O E$ from the center $O$ of the inscribed circle to the side $B C$. Then $O D=O E$. Let $A D$ $=D C=a, A B=B C=b, E F=x$.

Since

$$
S_{\triangle \mathrm{AOD}}=\frac{1}{2} A D \cdot O D, S_{\triangle \mathrm{OFB}}=\frac{1}{2} B F \cdot O E, S_{\triangle \mathrm{AOD}}=\frac{3}{8} S_{\triangle \mathrm{OFB}}
$$

then

$$
B F=\frac{8}{3} a=b-a-x
$$

By the property of the bisector of a triangle,

$$
\frac{A C}{F C}=\frac{A B}{B F}, \text { or } \frac{2 a}{a+x}=\frac{b}{b-a-x} .
$$

Eliminating $x$ from the obtained equations, we get that $\frac{a}{b}=\frac{1}{4}$. Therefore,

$$
\frac{A C}{A B}=\frac{2 a}{b}=\frac{1}{2}
$$

## Answer

$\frac{1}{2}$.