For example. (Beauty Giant: 26th Y Students Mathematics Competition Questions) In $\triangle A B C$, $\angle A<\angle C<90^{\circ}<\angle B$, construct the external angle bisectors of $\angle A$ and $\angle B$. If the lengths of these two external angle bisectors (from the vertex to the intersection point with the extended opposite side) are both equal to $A B$, find $\angle A$.


As shown in Figure 10, let $A D$ and $B E$ be the external angle bisectors of $\angle A$ and $\angle B$ respectively, with $A D = A B = B E$. Let $\angle A = \alpha$. Then, from $B E = A B$, we get $\angle B E A = \alpha$, so $\angle X B E = 2 \alpha$, and $\angle D B A = \angle X B C = 2 \angle X B E = 4 \alpha$. Also, from $A D = A B$, we get $\angle A D B = \angle A B D = 4 \alpha$.

And $\angle B A D = \frac{1}{2}(180^{\circ} - \angle A) = 90^{\circ} - \frac{1}{2} \alpha$,
so $\left(90^{\circ} - \frac{1}{2} \alpha\right) + 4 \alpha + 4 \alpha = 180^{\circ}$.

From this, we get $\alpha = 12^{\circ}$, i.e., $\angle A = 12^{\circ}$.