Solution:
Let $A B C D$ be a parallelogram. To show that $A B C D$ is a rectangle, we have to prove that one of its interior angles is $90^{\circ}$.
In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DCB}$
$A B=D C$ (Opposite sides of a parallelogram are equal)
$\mathrm{BC}=\mathrm{BC}$ (Common)
$\mathrm{AC}=\mathrm{DB}$ (Given)
$\therefore \triangle \mathrm{ABC} \cong \triangle \mathrm{DCB}$ (By SSS Congruence rule)
$\Rightarrow \angle \mathrm{ABC}=\angle \mathrm{DCB}$
It is known that the sum of the measures of angles on the same side of transversal is $180^{\circ}$.
$\angle \mathrm{ABC}+\angle \mathrm{DCB}=180^{\circ}(\mathrm{AB} \| \mathrm{CD})$
$\Rightarrow \angle \mathrm{ABC}+\angle \mathrm{ABC}=180^{\circ}$
$\Rightarrow 2 \angle \mathrm{ABC}=180^{\circ}$
$\Rightarrow \angle \mathrm{ABC}=90^{\circ}$
Since $A B C D$ is a parallelogram and one of its interior angles is $90^{\circ}, A B C D$ is a rectangle.
Let $A B C D$ be a parallelogram. To show that $A B C D$ is a rectangle, we have to prove that one of its interior angles is $90^{\circ}$.
In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DCB}$
$A B=D C$ (Opposite sides of a parallelogram are equal)
$\mathrm{BC}=\mathrm{BC}$ (Common)
$\mathrm{AC}=\mathrm{DB}$ (Given)
$\therefore \triangle \mathrm{ABC} \cong \triangle \mathrm{DCB}$ (By SSS Congruence rule)
$\Rightarrow \angle \mathrm{ABC}=\angle \mathrm{DCB}$
It is known that the sum of the measures of angles on the same side of transversal is $180^{\circ}$.
$\angle \mathrm{ABC}+\angle \mathrm{DCB}=180^{\circ}(\mathrm{AB} \| \mathrm{CD})$
$\Rightarrow \angle \mathrm{ABC}+\angle \mathrm{ABC}=180^{\circ}$
$\Rightarrow 2 \angle \mathrm{ABC}=180^{\circ}$
$\Rightarrow \angle \mathrm{ABC}=90^{\circ}$
Since $A B C D$ is a parallelogram and one of its interior angles is $90^{\circ}, A B C D$ is a rectangle.
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