Prove that if a parallelogram has equal diagonals, then it is a rectangle.

Suppose this parallelogram is ABCD, AC and BD are its diagonals, O is the intersection point of AC and DB.
AC = BD by condition; AO = CO, BO = DO (by the property of the diagonals and the point of their intersection in the parallelogram) ⇒ AO = BO = OC = OD. Then △ BOA = △ BOC = △ COD = △ AOD by 1 attribute, therefore, △ CDB = △ ABD = △ ABC = △ ADC and ∠ABC = ∠BCD = ∠CDA = BAD as the corresponding elements of equal figures. Whence each of these angles is 360 ° / 4 = 90 °.