Starting with a post on Facebook:
https://www.facebook.com/groups/646056952197053/permalink/2475000499302680/
In translation (thanks to Miss Google):
"Given the square ABCD and a point E on the side (BC). AF is the bisector of the angle \angle DAE,\;\;F \in (DC). Prove that AE=DF+BE."
Solution CiP
Let's rotate clockwise the square around the point A by 90^\circ degree.
Point D turns into D'=B. Point C turns into C' on the extension of the half-line [CB and point F will be in the new position F' \in (D'C')=(BC') so that BF'=DF.
If we denote the measure of the angle \angle BAE=2x^\circ, then
\angle EAF=\angle FAD=45^\circ-x^\circ.
Then \angle AFD=45^\circ+x^\circ and so \angle AF'D'=\angle AF'E=45^\circ+x^\circ.
This rotation allows us to say again \angle D'AF'=\angle BAF'=\angle DAF=45^\circ-x^\circ and then it's easy to see that
\angle F'AE=\angle F'AB+\angle BAE=(45^\circ-x^\circ)+2x^\circ=45^\circ+x^\circ.
We obtained that the triangle \Delta AEF' has equal angles \angle AF'E and \angle EAF', so it is an isosceles triangle with EA=EF'.
Now it turns out immediately
AE=EF'=EB+BF'=EB+D'F'=EB+DF.
\blacksquare
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A very interesting solution that uses relations between the areas of some triangles can be found here
https://www.facebook.com/photo.php?fbid=7507678955923909&set=p.7507678955923909&type=3
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Related to this problem we mention Problems #1334, #1363, #1366 from http://www.gogeometry.com/index.html