Question

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

Answer 1

I provide here diagram....

Step-by-step explanation:

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.

hope this will help you.....

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.