Question

How to prove that the parallelogram is a rectangle?

Answer 1

The diagonals are given equal.

So AC=BD....(1)

In triangles ABD and ABC,

BD=AC[From (1)]

AD=BC[opposite sides of a parallelogram are equal]

AB=AB[common]

So ABD and ABC are congruent by SSS axiom.

So angle DAB and angle CBA are equal by c.p.c.t

But DAB+CBA=180degrees since sum of adjacent angles=180degree

So DAB+DAB=180

2DAB=180

DAB=90

CBA=DAB=90

DCB=90 since they are opposite angles of parallelogram.

angle ADC is also 90 because it is=360-(90+90+90)[by angle sum property]

Hence all angles are 90.

So it is a rectangle.

Hope it helps.

Answer 2

Answer:

ABCD is a cyclic parallelogram.

To prove,

ABCD is a rectangle.

Proof:

∠1+∠2=180°      ...Opposite angles of a cyclic parallelogram

Also, Opposite angles of a cyclic parallelogram are equal.

Thus,

∠1=∠2

⇒∠1+∠1=180°

⇒∠1=90°

One of the interior angle of the parallelogram is right angled. Thus,

ABCD is a rectangle.

Step-by-step explanation: mark brainliest