Question

Prove that if the diagonals of a parallelogram are equal, then it is a rectangle. [4 MARKS]

Answer 1

Concept : 1 Mark
Application : 1 Mark
Proof : 2 Marks



Let $ABCD$ be the parallelogram with diagonal $AC$ = diagonal $BD$. Let O be their point of intersection.

We know that diagonals bisect each other in a parallelogram. It is also given that these diagonals are equal.

$\Rightarrow AO=BO=CO=DO.$

Let us assume that $\angle ADO=x$ and $\angle CDO=y$

$\Rightarrow \angle DAO=x$ and $\angle DCO=y$
[Since $\Delta COD,\Delta DOA$ are isosceles.]

Also, $\angle DAO=\angle BCO=x$ and $\angle DCO=\angle BAO=y$
[Pair of alternate angles]

$\Rightarrow \angle CBO=x$ and $\angle ABO=y$
[Since $\Delta COB,\Delta BOA$ are isosceles.]

$\angle A+\angle B+\angle C+\angle D={360}^{\circ }$ [By angle sum property]

i.e., $4x+4y={360}^{\circ }$

$\Rightarrow x+y={90}^{\circ }$.

Thus, the given parallelogram is a rectangle.