Example 3 Given one side, the angle opposite to it, and the angle bisector of this angle, construct a triangle.
Given: angle $\alpha$ and two segments $a, t_{a}$.
To construct: $\triangle A B C$, such that $B C=a$, $\angle A=\alpha$, and the angle bisector $A T=t_{a}$.
(1) Construct $\triangle P B C$, such that $B C=a, \angle P B C=\angle P C B=\frac{1}{2} \alpha$;
(2) Draw $B E \perp B P$ through $B$ and make $B E=\frac{1}{2} t_{a}$, connect $E P$, and on $E P$ intercept $E F=E B=$ $\frac{1}{2} t_{a}$
(3) With $P$ as the center and $P F$ as the radius, draw an arc intersecting $B C$ at $T$;
(4) Connect $P T$ and extend it to $A$, such that $T A=t_{a}$, connect $A B, A C$, then $\triangle A B C$ is the required triangle.
Proof (omitted)
Discussion The existence of a solution to this problem depends on whether point $T$ exists. If a perpendicular $P D$ is drawn from $P$ to $B C$ at $D$, then it depends on whether $P T$ is greater than $P D$. From

we have

we have
$$
\begin{array}{c}
P D=\frac{1}{2} a \cdot \tan \frac{\alpha}{2}, l=\frac{1}{2} a \cdot \sec \frac{\alpha}{2} \\
P F=\frac{1}{2} \sqrt{t_{a}^{2}+a^{2} \cdot \sec ^{2} \frac{\alpha}{2}}-\frac{1}{2} t_{a} \\
P F-P D=\frac{a t_{a}\left(\frac{a}{2 t_{a}}-\tan \frac{\alpha}{2}\right)}{\sqrt{t_{a}^{2}+a^{2} \cdot \sec ^{2} \frac{\alpha}{2}+a \cdot \tan \frac{\alpha}{2}}}
\end{array}
$$

Note that $\frac{\alpha}{2}$ must be an acute angle, hence when $\tan \frac{\alpha}{2} \leqslant \frac{a}{2 t_{a}}$, this problem has one solution (even though two positions of $T$ can be made, the two resulting triangles are congruent); when $\tan \frac{\alpha}{2}>\frac{a}{2 t_{a}}$, this problem has no solution.