Question

prove that if the diagonals of a parallelogram are equal,then its a rectangle

Answer 1

Solution:-
Let ABCD is a parallelogram whose sides are AB, BC, CD and AD.
In this,
AC = BD (its two diagonals)
Proof : In Δ ACB and Δ BDA, 
AC = BD (Given)
AB = BA (Common)
BC = AD (Opposite sides of the parallelogram ABCD)
Δ ACB ≡ Δ BDA (By SSS congruence)
∴ ∠ ABC = ∠ BAD .....(1) (CPCT)
Again, AD || BC  (Opposite sides of the parallelogram ABCD)
AD || BC and the transversal 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°
∴ ∠ = 90° and ∠ C = 90°
Similarly, we can prove that ∠ B = 90° and ∠ D = 90°
Therefore, the parallelogram ABCD is a rectangle.
Hence proved.