(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.

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 !