Awesome geometry - 9

There is no need to say angle CED is 63.43. It can be found out as follows. Let S be the side of the square.
$\tan CED =S/(S/2) = 2 \implies ~ CED = arctan(2)=63.43.\\~~\\~~\\$
$In~ \triangle EFA ,~\angle EFA = 90 - \angle FEA = complimentary ~angle~\angle CED.\\\implies~two~angles~of ~right~angled~\triangle s~EFA~and ~ CED~ are~ equal.\\\therefore~\triangle s~EFA ~ and ~ CED~are~\color{#D61F06}{SIMILAR}.\\\therefore~~in~\triangle s~~EFA~and~CED,~\dfrac{ hypotenuse ~ FE}{ hypotenuse ~CE}\\= \dfrac{EA}{CD}= \dfrac{S/2}{S}=1/2=\tan(26.57).\\But~in~ \triangle~FCE,~ \angle ECF~=~tan^{-1}~( \dfrac{FE}{CE} )=tan^{-1}(1/2)\\=\boxed{ \color{#20A900}{ 26.57}}$
Sorry I have made changes, it was to make it better.

It is easy to realize that two triangles CEF and CED are uniform. So: $\angle{CED}=\angle{EFC} \Rightarrow \angle{ECF}=90^{0}-\angle{EFC}=\boxed{26.57}$

We know that $\angle CDE = \angle EAF = 90^\circ$ and also $\angle DEC + \angle AEF = \angle DEC + \angle ECD = 180-90 = 90^\circ$

$\implies \angle AEF = \angle ECD \implies \triangle AEF \sim \triangle DEC$ (AA)

$\implies \angle EFA = \angle DEC = 63.43^\circ$

Now, prolong $\overline{CD}$ until it meets $f$ at $H$

We have $\angle AEF = \angle HED$ , $\angle EAF = \angle EDH = 90^\circ$ and, since $E$ is the midpoint of $\overline{AD}$ , $\overline{AE} = \overline{ED} \implies \triangle AEF \cong \triangle HED$ (ASA)

Now it gets interesting: from the congruence between $\triangle AEF$ and $\triangle HED$ and from the fact that $\angle CEF = 90^\circ$ , we can say that $\overline{EC}$ is both a median and a height of $\triangle FCH$ , which means that $\triangle FCH$ is isosceles, with $\angle CFE = \angle CHE = \angle EFA = \angle DEC = 63.43^\circ$ , and that $\overline{EC}$ is also the bisector of $\angle HCF$

$\implies \angle ECF = \dfrac{\angle HCF}{2} = \dfrac{180-2 \times 63.43}{2} = \boxed{26.57^\circ}$