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

Answer 1

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

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asked Dec 22, 2017 in Class IX Maths by saurav24 Expert (1.4k points)

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

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answered Dec 22, 2017 by navnit40 (-4,940 points)

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.

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Answer 2

$\huge\bf{ \underline{ \underline{Given :-}}}$

ABCD is a ||gm

therefore, i.e. AB = CD , AB||CD

AD = BC , AD || BC

$\huge\bf{ \underline{ \underline{To\: prove :-}}}$

ABCD is a rectangle

∠A = ∠B = ∠C = ∠D = 90°

$\huge\bf{ \underline{ \underline{Proof -:}}}$

In △ACD and △BCD

AD = BC (given)

CD = CD ( Common)

AC = BD (given)

therefore, △ACD ≅ △BCD (S.S.S)

therefore, ∠D = ∠C (C.P.CT)

Also AD || BC

∠D + ∠C = 180° (Co - interior ∠'s)

∠C + ∠C = 180°

2∠C = 180°

∠C = $\frac{180}{2}$ = 90°

|| ly we can prove that ∠A = ∠B = ∠D = 90°

therefore , ABCD is rectangle