Known: parallelogram ABCD, diagonal AC.BD The intersection point O, AEC and bed are all equal to 90 degrees Point E is on ad. a, B, C and D are all connected with E E is on the top of AD, sorry

Known: parallelogram ABCD, diagonal AC.BD The intersection point O, AEC and bed are all equal to 90 degrees Point E is on ad. a, B, C and D are all connected with E E is on the top of AD, sorry

It is proved that: the diagonals of parallelograms are equally divided with each other, and because OE is the center line of the hypotenuse of two RT △ ace and RT △ BDE, OB = od = OE = OA = OC, so a, B, C, D and E are all on the circle with o as the center and OE as the radius. So ∠ DAB + ∠ DCB = 180 ° and because the parallelograms are diagonally equal, ∠ DAB = 90 ° so the parallelograms ABCD are rectangular