Prove that if the diagonals of a quadrilateral are perpendicular, then the midpoints of its sides are the vertices

Prove that if the diagonals of a quadrilateral are perpendicular, then the midpoints of its sides are the vertices of the rectangle.

Consider triangles ABC and MBN. Angle B is common for them. Since AM = MB, AB = 2MB, then AB: MB = 2: 1. Similarly, we find that BC: BN = 2: 1. If the two sides of one triangle are proportional to the two sides of the other triangle, and the angles between these sides are equal, then such triangles are similar. AC: MN = 2: 1. Since ∆ABC and ∆MBN are similar, their corresponding angles are equal, i.e. the angles BAC and BMN are equal. For two straight lines AC, MN and a secant AB, the angles BAC and BMN are corresponding, which means that AC is parallel to MN. Similarly, considering ∆ADC and ∆PDK, we prove that AC: PK = 2: 1 and PK || AC. Since MN || AC and PK || AC, then MN || PK. AC: MN = 2: 1 and AC: PK = 2: 1, so MN = PK. If two opposite sides of a quadrilateral are equal and parallel, then such a quadrilateral is a parallelogram. MNKP is a parallelogram. Since AC is perpendicular to BD and MN || AC, MN is perpendicular to BD. Consider ∆AMF and ∆MBH. They are equal on the side and two corners adjacent to it. Angles AFM and MHB are equal and right. Consider the quadrangle FMHO. A quadrilateral is a parallelogram if its opposite angles are equal in pairs. And since we also have straight lines, then the FMHO quadrangle is a rectangle. In the quadrilateral MNKP, the angle FMH is right, MN || PK and MN = PK, so MNKP is a rectangle. Ch.t.d.