Isosceles Twins, part 1

Since $\angle AED = \angle DEC = 90°$ , $ED = ED$ , and the incircles of $\triangle AED$ and $\triangle DEC$ are congruent, then $\triangle AED \cong \triangle DEC$ .

Since $\triangle AED \cong \triangle DEC$ , $\angle EDA = \angle EDC$ , and since $\angle ADC = 90°$ , $\angle EDA = \angle EDC = 45°$ , so $\triangle AED$ and $\triangle DEC$ are isosceles right triangles, and $\angle EAD = \angle ECD = 45°$ .

The base angles of isosceles $\triangle ABC$ are then $\angle CAB = \angle CBA = \frac{1}{2}(180° - 45°) = 67.5°$ , so $\angle DAB = \angle CAB - \angle EAD = 67.5° - 45° = 22.5°$ .

That makes $AD = AB \cos 22.5° = \cfrac{\sqrt{2 + \sqrt{2}}}{2}$ , and $AE = ED = \cfrac{1}{\sqrt{2}} AD = \cfrac{\sqrt{2 + \sqrt{2}}}{2\sqrt{2}}$ .

Therefore, the inradius $r = \cfrac{1}{2}(AE + ED - AD) = \cfrac{1}{2}\bigg(\cfrac{\sqrt{2 + \sqrt{2}}}{2\sqrt{2}} + \cfrac{\sqrt{2 + \sqrt{2}}}{2\sqrt{2}} - \cfrac{\sqrt{2 + \sqrt{2}}}{2}\bigg) = \cfrac{\sqrt{2 - \sqrt{2}}}{4}$ , so $a = 2$ , $b = 4$ , and $a + b = \boxed{6}$ .