12.173. In triangle $ABC$, angle $A$ is equal to $\alpha$ and side $BC=a$. Find the length of the bisector $AD$, if the angle between the bisector $AD$ and the altitude $AE$ is $\beta$.
## Solution.

If $A D=x$ (Fig. 12.43), then from $\triangle A E D\left(\angle A E D=90^{\circ}\right)$:

$A E=A D \cos \angle E A D=x \cos \beta$, and from $\triangle A E B\left(\angle A E B=90^{\circ}\right)$:

$$
B E=A E \operatorname{tg} \angle B A E=x \cos \beta \operatorname{tg}\left(\frac{\alpha}{2}-\beta\right)
$$

In $\triangle A E C\left(\angle A E C=90^{\circ}\right): E C=A E \operatorname{tg} \angle E A C=x \cos \beta \operatorname{tg}\left(\frac{\alpha}{2}+\beta\right)$.

Since $B C=B E+E C$, then:

$$
\begin{aligned}
& a=x \cos \beta\left(\operatorname{tg}\left(\frac{\alpha}{2}-\beta\right)+\operatorname{tg}\left(\frac{\alpha}{2}+\beta\right)\right)=\frac{x \cos \beta \sin \alpha}{\cos \left(\frac{\alpha}{2}-\beta\right) \cos \left(\frac{\alpha}{2}+\beta\right)} \Leftrightarrow x= \\
& =\frac{a \cos \left(\frac{\alpha}{2}-\beta\right) \cos \left(\frac{\alpha}{2}+\beta\right)}{\cos \beta \sin \alpha}
\end{aligned}
$$

$$
\text { Answer: } \frac{a \cos \left(\frac{\alpha}{2}-\beta\right) \cos \left(\frac{\alpha}{2}+\beta\right)}{\cos \beta \sin \alpha}
$$