Theorem: The quadrisectors that divide the angles of a rhombus ABCD into four equal parts are drawn. For each pair of adjacent angles, those quadrisectors that are closest to the enclosed side are extended until a point of intersection is established. The line segments connecting those points of intersection form a square EFGH.
Proof: Refer to the associated dynamic sketch: By symmetry of the rhombus (symmetrical around its diagonals) and the symmetrical placement of points, EFGH is a rectangle. But E is the incentre of right triangle ABM, since AE and BE are angle bisectors by construction. Hence, EM bisects right angle BMA. Therefore, the diagonals EG and FH of rectangle EFGH are perpendicular to each other and it follows that EFGH is a square.