Question

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

Answer 1

Answer:

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this is your answer

Given: 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.

hope it helped you!!

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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.