Question

Prove that each angle of a rectangle is ${90}^{\circ }$.

Answer 1

We know that opposite sides of a rectangle are equal. Therefore, $AD=BC$ and $AB=CD$. Also, the diagonals of rectangle are also equal. Therefore, $AC=BD$

Now, consider $\Delta ADC$ and $\Delta BCD$

$\Rightarrow AD=BC$

$\Rightarrow AC=BD$

$\Rightarrow DC=DC$ [Common]

$\therefore \Delta ADC\cong \Delta BCD$ by SSS congruency rule.

Therefore, $\angle ADC=\angle BCD$ [By CPCT]

But, adjacent sides of a parallelogram are supplementary. [Since, rectangle is a parallelogram].

$\therefore \angle ADC+\angle BCD={180}^{\circ }$

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

Since, opposite angles of a parallelogram are also equal.

Therefore, we can say that each angle of a rectangle is ${90}^{\circ }$.