Hello, JacobFischer!
Let $\displaystyle \theta$ = angle A.
We will calculate all the angles in the diagram.
Write these figures in your diagram as we go along.
Since $\displaystyle \Delta ADF$ is isosceles, $\displaystyle \angle DFA = \theta.$
Since $\displaystyle \angle EDF$ is an exterior angle, $\displaystyle \angle EDF = 2\theta.$
Since $\displaystyle \Delta EDF$ is isosceles, $\displaystyle \angle DEF = 2\theta.$
Then $\displaystyle \angle DFE \:=\:180^o - 2\theta - 2\theta \:=\:180^o - 4\theta$
Hence, $\displaystyle \angle EFC \:=\:180^o - \theta - (180^o-\theta)\:=\:3\theta$
Since $\displaystyle \Delta FEC$ is isosceles, $\displaystyle \angle ECF = 3\theta.$
Then $\displaystyle \angle FEC \:=\:180^o - 3\theta - 3\theta \:=\:180^o - 6\theta$
Hence, $\displaystyle \angle BEC \:=\:180^o - 3\theta - (180^o - 6\theta) \:=\:4\theta$
Since $\displaystyle \Delta ECB$ is isosceles, $\displaystyle \angle EBC = 4\theta$
Hence, $\displaystyle \angle ECB \:=\:180^o - 4\theta - 4\theta \:=\:180^o - 8\theta$ .[1]
But we know that: $\displaystyle \angle ECB + \angle ECF \:=\:90^o$
. . That is: .$\displaystyle \angle ECB + 3\theta \:=\:90^o$ .[2]
Substitute [1] into [2]: .$\displaystyle (180^o - 8\theta) + 3\theta \:=\:90^o $
. . and we have: .$\displaystyle \text{-}5\theta \:=\:\text{-}90^o \quad\Rightarrow\quad \boxed{\theta \:=\:18^o}$