Question

if the diagonals of parallelogram are equal then prove that it is 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 90°

In ΔABC and ΔDCB,

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

BC = BC (Common)

AC = DB (Given that the diagonals are equal)

ΔABC ΔDCB (By SSS Congruence rule)

∠ABC = ∠DCB

It is known that the sum of the measures of angles on the same side of transversal is 180°

∠ABC + ∠DCB = 180°(AB || CD)

∠ABC = 90°

Since ABCD is a parallelogram and one of its interior angles is 90°, ABCD is a rectangle.

Hope it helps you ❣️☑️☑️

Answer 2

Given : A parallelogram ABCD , in which AC = BD

TO Prove : ABCD is a rectangle .

Proof : In △ABC and △ABD

AB = AB [common]

AC = BD [given]

BC = AD [opp . sides of a | | gm]

⇒ △ABC ≅ △BAD [ by SSS congruence axiom]

⇒ ∠ABC = △BAD [c.p.c.t.]

Also, ∠ABC + ∠BAD = 180° [co - interior angles]

⇒ ∠ABC + ∠ABC = 180° [∵ ∠ABC = ∠BAD]

⇒ 2∠ABC = 180°

⇒ ∠ABC = 1 /2 × 180° = 90°

Hence, parallelogram ABCD is a rectangle.