Dealing with angles (1)

Let $M$ be the intersection of $DF$ and $BE$ . Since $DF\mid\mid AC$ , we know

$\angle FDE=\angle ACE=\alpha$

$\angle FMB=\angle MBC=\beta$

Then

$A=a°+\alpha -\beta-90°$

$B=90°-\beta-(\angle AEC)$

By adding $\beta$ and using $\angle AFC=180°-\alpha-a°$ :

$A=a°+\alpha-90°$

$B=90°-(180°-\alpha-a°)=\alpha+a°-90°$

Therefore $\boxed{A=B}$ .

Let $DF$ and $BE$ intersect at $P$ . Note that opposite $\angle DPE = \angle BPF = b$ . And $b+d+e = 180$ , $\implies \color{#3D99F6} d = 180 - e=b$ . We also note that external $\angle CBE = \angle BAE + \angle AEB$ , $\implies c = a+f$ . $\implies \color{#D61F06} a-c = - f$ . Then we have:

$\begin{aligned} A & = {\color{#D61F06}a} + {\color{#3D99F6}d} \ {\color{#D61F06}-c} - 90 \\ & = {\color{#3D99F6} 180 -e-b} \ {\color{#D61F06}-f} - 90 \\ & = 90-e-b-f \\ & = B \end{aligned}$

$\implies \boxed{A=B}$ . Note that the answer is also true for $AC$ not parallel to $FD$ .