43. In triangle $ABC$, all sides of which are different, the bisector of angle $A$ intersects side $BC$ at point $D$. It is known that $|AB|-|BD|=a, |AC|+$ $+|CD|=b$. Find $|AD|$.
43. Let's take a point $K$ on side $A B$ such that $|B K| \neq |B D|$, and a point $E$ on the extension of $A C$ such that $|C E|=|C D|$. We claim that $\triangle A D K$ is similar to $\triangle A D E$. If $\hat{A}, \hat{B}$, and $\hat{C}$ are the measures of the angles of $\triangle A B C$, then

$$
\begin{aligned}
& \widehat{D K A}=180^{\circ}-\widehat{D K B}=180^{\circ}-\left(90^{\circ}-\frac{\hat{B}}{2}\right)=90^{\circ}+\frac{\hat{B}}{2} \\
& \widehat{A D E}=180^{\circ}-\widehat{C E D}-\frac{\hat{A}}{2}=180^{\circ}-\frac{\hat{A}}{2}-\frac{\hat{C}}{2}=90^{\circ}+\frac{\hat{B}}{2}
\end{aligned}
$$

Thus, $\widehat{A K D}=\widehat{A D E}$. Additionally, by the condition $\widehat{D A E}=$ $=\widehat{D A K}$.

Answer: $|A D|=\sqrt{\overline{a b}}$.