Answer: 547

Small Hint:

Let $\angle BAC = \alpha$ and chase angles along the chain of isosceles triangles $AED,$ $EDB,$ and $DBC.$

Big Hint:

The exterior angle gives $\angle DEB = 2\alpha;$ then the three angles at $D$ along $\overline{AC}$ force $\angle BDC = 3\alpha = \angle BCD.$

Solution:

Let $\angle BAC = \alpha.$ Since $AE = ED,$ triangle $AED$ is isosceles with $\angle ADE = \angle DAE = \alpha,$ so the exterior angle at $E$ gives $\angle DEB = 2\alpha.$ Since $ED = DB,$ triangle $EDB$ has $\angle DBE = \angle DEB = 2\alpha,$ hence $\angle EDB = 180^\circ - 4\alpha.$

The three angles at $D$ on segment $\overline{AC}$ sum to a straight angle: $\alpha + (180^\circ - 4\alpha)$ $+ \angle BDC = 180^\circ,$ so $\angle BDC = 3\alpha.$ Since $DB = BC,$ also $\angle BCD = \angle BDC = 3\alpha.$ But $AB = AC$ makes $\angle ABC = \angle ACB = 3\alpha,$ so the angle sum of $\triangle ABC$ gives $\alpha + 3\alpha + 3\alpha = 180^\circ,$ hence $\alpha = \frac{180}{7}$ degrees.

Then $\angle ABC = 3\alpha = \frac{540}{7}$ degrees, and $m + n = 540 + 7 = 547.$