In triangle $A B C$, it is known that $A B=c, A C=b$, and the bisector coming out of angle $A$ is equal to $l$. Find the third side of the triangle.
Let $A D=l$ be the bisector of triangle $A B C, B D=c^{\prime}, C D=b^{\prime}$. First, we will prove that the square of the bisector of a triangle is equal to the product of the sides enclosing it, minus the product of the segments of the third side, into which it is divided by the bisector, i.e., $l^2=b c-b^{\prime} c^{\prime}$. Let $M$ be the point of intersection of the extension of the bisector $A D$ of triangle $A B C$ with the circumcircle of this triangle. Then triangle $A B D$ is similar to triangle $A M C$ by two angles. Therefore,

$$
\begin{aligned}
& \frac{A D}{A C}=\frac{A B}{A M}, \text{ so } A D(A D+D M)=A C \cdot A B \\
& l(l+D M)=b c, l^2=b c-l \cdot D M=b c-b^{\prime} c^{\prime}
\end{aligned}
$$

($l \cdot D M=b^{\prime} c^{\prime}$ by the theorem on the products of segments of intersecting chords), which is what we needed to prove. By the property of the bisector of a triangle $\frac{f^{\prime}}{\frac{1}{Z}}=\frac{B D}{C D}=\frac{\varepsilon}{b}$. Let $c^{\prime}=c x, b^{\prime}=b x$. Then $l^2=b c-b^{\prime} c^{\prime}=b c-b c x^2$, from which $x=$ $\sqrt{\frac{b x-l^2}{b x}}$.

Therefore,

$$
B C=b^{\prime}+c^{\prime}=(b+c) x=(b+c) \sqrt{\frac{\left(b x-l^{2}\right)}{b r^{2}}} .
$$

## Answer

$(b+c) \sqrt{\frac{b x-l^{2}}{b r}}$.