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.
