### Proof

After to make use a square on the other, we trace segment BE and perpendicular straight lines r and s to the EB passing, respectively, for points and and B. Are H the point of r on segment FG. We trace the perpendicular straight line to the EH passing for H and either I the point of intersecção of this straight line with s. perpendicular Traçamos IJ to the BG.
After this construction we have that BEHI is rectangular. We go then, initially, to show that of fact it is squared, for this showing that EB and EH are congruentes.Os triangles EFH and ECB are congruentes since both rectangles, with congruentes legs EF and EC (are sides of square ECGF) and congruentes angles FEH and CEB (each one of them added to complete angle CEH 90 degrees).
 On the congruence that guarantees the rabbet of the parts: *triangles FEH and JBC are congruentes therefore rectangular and have congruentes sides EH and BC (opposing sides of rectangle), congruentes angles FEH and JBI (angles with parallel sides) *triangles LAB and MJI are congruentes therefore rectangular with congruentes sides AB and JI (congruentes AB and BC, and BC in turn congruente to the JI, since for transitividade triangles JBI and CEB are congruentes) and congruentes angles ABL and JIM (angles with parallel sides)
*triangles HGM and EDL are congruentes therefore rectangular with congruentes sides EL and HM (EL=EB-LB, HM=HI-MI, and are congruentes EB and HI, as well as LB and MI) and congruentes angles DEL and GHM (angles with parallel sides).

The verification of the congruences in guarantees them that the clippings made in squares ABCD and ECGF, in fact, allow to mount square biggest BEHI.

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