 Hello and welcome to the session. I am Asha and I am going to help you with the following question which says, if the diagonals of a parallelogram are equal, then show that it is a rectangle. Let us begin with the solution and let ABCD be the given parallelogram. So we are given a parallelogram in which the diagonals are equal, that is AC is equal to BD. And we have to show that parallelogram ABCD is a rectangle. Now in triangle ADC and triangle BCD, we have DC is equal to CD. This is common to both the triangles. Also AC is equal to BD since this is given to us the diagonals of the parallelogram are equal and AD is equal to BC since opposite sides of a parallelogram are equal. Therefore by SSS congruence, we can say that triangle ADC is common to triangle BCD which further implies that angle ADC is equal to angle BCD. That is these two angles are equal. Let this be equation number one. Now in a parallelogram opposite sides are parallel. So AD is parallel to BC and the transversal DC intersects them. So this implies that angle ADC plus angle BCD is equal to 180 degree. Since if two parallel lines are intersected by a transversal, then sum of consecutive interior angles is supplementary. That is 180 degree. And now since angle ADC is equal to angle BCD, therefore this step can further threaten us two times of angle ADC is equal to 180 degree. Or we can say that angle ADC is equal to 180 degree divided by 2 which is equal to 90 degree. Thus angle ADC is equal to 90 degree. Now from equation one, angle ADC is equal to angle BCD. So this implies angle BCD is also equal to 90 degree. Similarly we can show that the other two angles that is angle DAB and angle CBA are also 90 degree. Angle DAB is equal to angle CBA is equal to 90 degree. Thus this implies all four angles of parallelogram ABCD 90 degree each. So this implies parallelogram ABCD is a rectangle. This completes the session. Take care and bye for now.