Given an irregular pentagon ABCDE with AB = AE = CD = BC+DE, and right angles in B and D.
(first of all construct such a pentagon...) may be BC ≠ DE.
Disect this pentagon into 3 pieces to build a square.
A,B,C draggable.

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The pentagon is constructed as follows :
From AB, draw the perpendicular in B to AB, choose any point C with BC < AB, then draw C' with CC' = AB, B ∈ CC'.
D is reflection of C' through AC, T reflection of B through AC, and E reflection of T through AD.

From the solution square ABB'A', it is even simpler :
Choose any C' on BB', draw AC' and A'C', cut and rearrange into the pentagon.
Note : triangles AA'C' and DCA are images of each other through a mirror. This piece then has to be flipped !


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