Question

If one of the angles of a parallelogram is a right angle, prove that it is a rectangle.

Answer 1

Let $ABCD$ is a parallelogram in which $\angle ABC={90}^{\circ }$

Since $ABCD$ is a parallelogram

$\angle ADC=\angle ABC$ (opposite angles of a parallelogram are equal)

$\angle ADC={90}^{\circ }$

Now, $\angle ABC+\angle BCD={180}^{\circ }$ (sum of co-interior angles)

$⟹{90}^{\circ }+\angle BCD={180}^{\circ }$

$⟹\angle BCD={90}^{\circ }$

$\angle DAB=\angle BCD$ (opposite angles of a parallelogram)

$⟹\angle DAB={90}^{\circ }$

Therefore, $\angle ABC=\angle BCD=\angle ADC=\angle DAB={90}^{\circ }$

Since $ABCD$ is a parallelogram in which each angle is of ${90}^{\circ }$

Hence, $ABCD$ is a rectangle.