Question

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

Answer 1

Answer:

□ 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:

Gven: In parallelogram ABCD, AC=BD

To prove : Parallelogram ABCD is rectangle.

Proof : in △ACB and △BDA

AC=BD ∣ Given

AB=BA ∣ Common

BC=AD ∣ Opposite sides of the parallelogram ABCD

△ACB ≅△BDA∣SSS Rule

∴∠ABC=∠BAD...(1) CPCT

Again AD ∥ ∣ Opposite sides of parallelogram ABCD

AD ∥BC and the traversal AB intersects them.

∴∠BAD+∠ABC=180∘

...(2) _ Sum of consecutive interior angles on the same side of the transversal is

180∘

From (1) and (2) ,

∠BAD=∠ABC=90∘

∴∠A=90∘

and ∠C=90∘

Parallelogram ABCD is a rectangle.

and ∠C=90∘

Parallelogram ABCD is a rectangle.