Example 19 (Problem 1506 from "Mathematics Bulletin") In $\triangle A B C$, $A B=A C$, the angle bisector of $\angle B$ intersects $A C$ at $D$, and $B C=B D+A D$. Find $\angle A$.

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Solution 1 As shown in Figure 6-3, take a point $E$ on $BC$ such that $BE = BD$. Connect $DE$. Since $AB = AC$, we have $\angle ABC = \angle C$.
Let $\angle C = 2\alpha$. Since $\angle ABD = \angle CBD$, we have $\angle ABD = \angle CBD = \alpha$.
Since $BC = BD + AD$, we have $BC = BE + CE = BD + CE$.
Thus, $AD = CE$.
By the Angle Bisector Theorem, we know $\frac{AB}{BC} = \frac{AD}{CD}$,
So $\frac{AC}{BC} = \frac{CE}{CD}$.
Since $\angle C = \angle C$, we have $\triangle ABC \sim \triangle EDC$.
Thus, $\angle CDE = \angle ABC = 2\alpha$.
Therefore, $\angle BED = \angle BDE = 2\alpha + 2\alpha = 4\alpha$.
Since in $\triangle BDE$, $\alpha + 4\alpha + 4\alpha = 180^\circ$,
We have $\alpha = 20^\circ$.
Thus, $\angle A = 180^\circ - 4\alpha = 100^\circ$.

Solution 2 As shown in Figure 6-4, construct the circumcircle $\odot O$ of $\triangle ABD$, and let it intersect $BC$ at $E$. Connect $DE$.

Take a point $M$ on $BC$ such that $BM = AB$.
Since $AB = AC$, we have $\angle ABD = \angle CBD$, so let $\angle ABD = \angle CBD = \alpha$.
Thus, $\angle ACB = 2\alpha$. By the property of a cyclic quadrilateral, we have $\angle CDE = \angle ABC = 2\alpha$. So $\angle CDE = \angle ECD = 2\alpha$, thus $DE = CE$.
Since $\triangle ABD \cong \triangle MBD$, we have $\angle MDB = \angle ADB = \alpha + 2\alpha = 3\alpha$.
Thus, $\angle DME = \angle CBD + \angle BDM = \alpha + 3\alpha = 4\alpha$.

And $\angle DEM = \angle C + \angle CDE = 2\alpha + 2\alpha = 4\alpha$.
So $\angle DME = \angle DEM$, $\angle EDC = \angle ECD$,
Thus, $AD = DM = DE = EC$.
Since $BC = BD + AD$, $BC = BE + CE$.
Thus, $BD = BE$, so $\angle BDE = \angle BED = 4\alpha$.
In $\triangle BDE$, $\alpha + 4\alpha + 4\alpha = 180^\circ$, so $\alpha = 20^\circ$.
Thus, $\angle A = 180^\circ - 4\alpha = 100^\circ$.