Question

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

Answer 1

Answer:

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.

Step-by-step explanation:

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Answer 2

Answer:

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.