Question

If the diagonals of a parallelogram are equal, then show that it is a rectangle.
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Answer 1

Answer:

Step-by-step explanation:

□ 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.

Answer 2

Step-by-step explanation:

Let the parallelogram be ABCD, with AC and BD as diagonals and AC=DB.

In △ABC and △DCB

=>AB=DC (Opposite sides of a ||gm are equal)

AC=DB (Given)

BC=BC (Common Side)

=>△ABC≅△DCB (By SAS criterion)

=>∠ABC=∠DBC

Since ABCD is a ||gm, the sum of adjacent two angles is 180°.

Or, we can also considet that co-interior angles formed by the transversal of any two opposite and parallel sides of ||gm sum 180°.

=>∠ABC+∠DCB=180°

But we know that ∠ABC=∠DBC.

=>∠ABC+∠ABC=180°

=>2×∠ABC=180°

=>∠ABC=90°

And, ∠DCB=90°

Similarly other two angles of this parallelogram are also 90° each.

Since all the angles of this parallelogram are 90° each, this parallelogram is a rectangle.