8,9 In an isosceles triangle $A B C$ ( $A B=B C$ ), a circle is inscribed. A tangent to the circle is drawn through point $M$, lying on side $A B$, intersecting line $A C$ at point $N$. Find the lateral side of triangle $A B C$ if $A C=C N=a, M B=\frac{1}{\bar{\varepsilon}} A B$.
Let the circle touch the lines $A B, A C$, and $M N$ at points $P, S$, and $T$ respectively. Denote $A B=x, \angle B A C=\alpha$. Then

![](https://cdn.mathpix.com/cropped/2024_05_06_3af6d4d9a845a74ce61dg-11.jpg?height=77&width=1140&top_left_y=156&top_left_x=459)

$$
\begin{aligned}
& N T=N S=N C+C S=a+\frac{1}{2} a=\frac{3}{2} a, M N=N T+M T=\frac{\frac{3}{2}}{2} a+\left(\frac{7}{8} x-\frac{1}{2} a\right)=a+\frac{7}{8} x, \\
& \cos \alpha=\cos \angle B A C=\frac{A S}{A B}=\frac{a}{2 x}
\end{aligned}
$$

By the Law of Cosines, $M N^2=A M^2+A N^2-2 A M \cdot A N \cos \alpha$,

$$
\begin{gathered}
\text { or } \\
\left(a+\frac{7}{8} x\right)^2=\left(\frac{7}{8} x\right)^2+4 a^2-2 \cdot \frac{7}{8} x \cdot 2 a \cdot \frac{a}{2 x}, \frac{7}{4} x=3 a-\frac{7}{4} a .
\end{gathered}
$$

From this, we find that $x=\frac{5}{7} a$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3af6d4d9a845a74ce61dg-11.jpg?height=580&width=1907&top_left_y=842&top_left_x=110)

## Answer

$\frac{5}{7} a$

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