Question

if the diagnols of parallellogram are equal then show that it is a rectangle
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Answer 1

Step-by-step explanation:

ANSWER

Ref.Image

□ ABCD is a parallelogram

consider Δ ACD and Δ ABD

AC = BD .... (given)

AB = DC .... (opposite sides of parallelogram)

AD = AD .... (common side)

∴Δ ACD ≅Δ ABD (sss test of congruence)

∠ BAD = ∠ CDA .... (cpct)

∠BAD+∠CDA=180

∘

. [Adjacent angles of parallelogram are supplementary]

so ∠ BAD and ∠ CDA are right angles as they are congruent and supplementary.

Therefor, □ ABCD is a rectangle since a

parallelogram with one right interior angle is a rectangle.

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Answer 2

Given:

A parallelogram ABCD with diagonals AC & BD.

To prove:

ABCD is a rectangle.

Proof:

In ∆ABC & ∆ABD

AB=AB ( common )

AC=BD (given )

BC=AD ( opposite 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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