Question

if the diagonals of parallelogram are equal ,then show that it is a rectangle.​

Answer 1

Answer:

Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that one of

its interior angles is 900.

In ΔABC and ΔDCB,

AB = DC (Opposite sides of a parallelogram are equal)

BC = BC (Common)

AC = DB (Given)

By SSS congruence rule,

ΔABC  ≅ ΔDCB

So, ∠ABC = ∠DCB

It is known that the sum of measures of angles on the same side of traversal is 1800

     ∠ABC + ∠DCB = 1800                  [AB || CD]

=> ∠ABC + ∠ABC = 1800

=> 2∠ABC = 1800

=> ∠ABC = 900

Since ABCD is a parallelogram and one of its interior angles is 900, ABCD is a rectangle.

FOLLOW  ME`

Answer 2

Step-by-step explanation:

In the photo,

Given:

• Abcd is parallogram,
• diagonals AC = BD

Prove :

• ABCD is rectangle

Proof:

In triangle ABC and DAB

AB = AB ( common)

BC= AD (ABCD is parallogram)

AC = BC ( given)

∆ABC =~ ∆BAD (SSS)

Angle a = Angle b (cpct(

Angle a + ANGLE b = 180

(adjacent angles of parallogram)

2 Angle a = 180

Angle a = 90°

since ABCD is parallogram with one Angle 90°, ABCD is rectangle