Question

prove that parallelogram is a square when the diagonals of a parallelogram are equal and intersect at right angle.

Answer 1

Let there be a parallelogram ABCD whose diagonals are equal and intersect at right angle at point O.

Now in trig. AOB

angle AOB = 90

AO = BO

Hence,

ang. ABO = ang. BAO = 45

similarly in all trig. have their base angle 90

& opposite sides are equal because it is a parallelogram.

Hence ABCD is a rectangle

Now in trig. AOB &  BOC we have,

OB = OB

ang.  AOB =  ang. BOC  = 90

AO = CO = 1/2 AC

Hence AB = BC

and a rectangle having adjacent side equal is a square hence

ABCD is a square

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Answer 2

Answer:

Let there be a parallelogram ABCD whose diagonals are equal and intersect at right angle at point O.

Now in trig. AOB

angle AOB = 90

AO = BO

Hence,

ang. ABO = ang. BAO = 45

similarly in all trig. have their base angle 90 & opposite sides are equal because it is a parallelogram.

Hence ABCD is a rectangle

Now in trig. AOB & BOC we have,

OB = OB

ang. AOB ang. BOC = 90

AO CO = 1/2 AC

Hence AB = BC

and a rectangle having adjacent side equal is a square hence

ABCD is a square

If you did not find it helpful talk to me at brainly