Question

The digonals of parallelogram are equal then show that it is a rectangle​

Answer 1

Question:

The digonals of parallelogram are equal then show that it is a rectangle.

Answer:

(Refer the attachment)

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 to both triangles)

➞AC = DB .........(Given)

By SSS congruence rule,

➞ΔABC ≅ ΔDCB

Therefore, ∠ABC = ∠DCB

We know that ,

The sum of measures of angles on the same side of traversal is 180°.

Therefore,

➞∠ABC + ∠DCB = 180° .............[ since , AB || CD]

➞∠ABC + ∠ABC = 180°

➞ 2 ( ∠ABC ) = 180°

➞ ∠ABC = 90°

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

ABCD is a rectangle.

Hence it is proved .

Answer 2

Step-by-step explanation:

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.