Prove that sum of angles is constant We are given square ABCD, point E on the extension of side CD and CF perpendicular to EB.
Find the value of angles $α+β$.

I am trying to prove this by triangles similarity.
I can see that $\angle \alpha$ is equal to $\angle \alpha 1$, which means that $\alpha+\beta$ is always $90^\circ$.
To prove this, it would suffice to show that triangles $ABE$ and $ABF$ are similar. They share an angle $\angle ABE$, so I must also show that $\angle AFB$ is equal to $\angle EAB$.
I have tried several angles relations but can't see anything that proves the required.
Any ideas please?
PS: The source of the problem is a Facebook group - it was given to me by a friend who can't solve it either.
Thank you in advance!
 A: Let the diagonals $AC$ and $BD$ intersect in the point $O$, which is the center of the square $ABCD$ of the square. Perform a $90^{\circ}$ counter-clockwise rotation around the point $O$.

Then, the line $CD$ is rotated to the line $DA$ and so the image $G$ of the point $E$ under this rotation is on the line $DA$. Because of the rotation, the segment $BE$ is rotated to the segment $CG$ and so $CG$ is perpendicular to $BE$. However, by assumption, the line $CF$ is also perpendicular to $BE$, so, by the uniqueness of perpendicularity, the point $F$ must lie on the segment $CG$ and so $\angle\, GFB = 90^{\circ}$. Again, because of the rotation, the triangles $\Delta\, ABE$ and $\Delta\, CBG$ are congruent and $$\angle\, BGF = \angle\, BGC = \angle\, AEB = \alpha$$
Since
$$\angle \, BAG = 90^{\circ} = \angle \, BFG$$ the quad $BFAG$ is cyclic and therefore
$$\angle\, BAF = \angle \, BGF = \alpha$$
Consequently
$$\alpha + \beta = \angle\, BAF + \angle\, FAD = \angle \, BAD = 90^{\circ}$$
A: $\triangle BFC$ and $\triangle ECB$ are similar, therefore $$\frac{BF}{BC}=\frac{BC}{BE}$$ Also $$AB=BC$$ hence,$$\frac{BF}{AB}=\frac{AB}{BE}$$ From SAS rule $\triangle AFB$ and $\triangle EAB$ are similar.
A: I know you want to prove it through relations of similar triangles, but solving the math for the angles might help you... (Look @endgame yourgame 's comment first):
Let the intersection of lines $AD$ and $BE$ = $P$
$\angle FPA = 180 - (360 - \angle CFE - \angle CDA - (\angle FCD))$, where $\angle FCD = (90 - \phi)$
$\angle FPA = 180 - (360 - 90 - 90 - (90 - \phi)) = 180 - (90 + \phi) = \mathbb{90 - \phi}$
$\angle BAD = \alpha + \beta = 180  - (90 - \phi) - \phi = 180 - 90 + \phi - \phi = \mathbb{90}$
$\therefore \alpha + \beta = 90˚$
