A Square Problem.

Since the triple $(21,20,29)$ is a primitive pythagorean triple $\implies \triangle{EFD}$ is a right triangle.

Let $m\angle{EDA} = \theta \implies m\angle{AED} = 90 - \theta$ and $m\angle{BEA} = 180 - 90 - (90 - \theta) = \theta \implies \triangle{AED} \sim \triangle{BEF}$

Let $AD = x$ and $AE = y \implies EB = x - y$ .

$\triangle{AED} \sim \triangle{BEF}\implies \dfrac{x}{21} = \dfrac{x - y}{20} \implies y = \dfrac{x}{21}$

Using right $\triangle{AED} \implies \dfrac{x^2}{21^2} + x^2 = 21^2 \implies (21^2 + 1)x^2 = 21^4$

$\implies x = \dfrac{21^2}{\sqrt{21^2 + 1}} = \dfrac{441}{\sqrt{442}}$ $\implies \alpha = |\overline {\rm FC}| = \sqrt{29^2 - \dfrac{441^2}{442}}$

$= \sqrt{\dfrac{177241}{442}} = \dfrac{421}{\sqrt{442}} = \dfrac{a}{\sqrt{b}} \implies a + b = \boxed{863}$ .