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

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.

2 x t divided by n

step-by-step explanation: