Question

Suppose the opposite angles of a quadrilateral are equal. Prove that it is a parallelogram.

Answer 1

$\Rightarrow$ In a quadrilateral ABCD,

$\Rightarrow$ $\angle A=\angle C$ and $\angle B=\angle D$ [Given] --- ( 1 )

$\Rightarrow$ We know that, sum of interior angles of quadrilateral is ${360}^{\circ }$.

$\Rightarrow$ $\angle A+\angle B+\angle C+\angle D={360}^{\circ }$

$\Rightarrow$ $\angle A+\angle B+\angle A+\angle B={360}^{\circ }$ [From ( 1 )]

$\therefore$ $\angle A+\angle B={180}^{\circ }$

$\Rightarrow$ Similarly, $\angle C+\angle D={180}^{\circ }$

$\Rightarrow$ Extend DC to E.

$\Rightarrow$ $\angle C+\angle BCE={180}^{\circ }$

$\Rightarrow$ Hence, $\angle BCE=\angle ADC$

$\Rightarrow$ If $\angle BCE=\angle D$ then,

$\Rightarrow$ $AD\parallel BC$ [$\because$ DC as a transversal] ---( 2 )

$\Rightarrow$ We can similarly show $AB\parallel DC$ -- ( 4 )

From ( 2 ) and ( 3 ) we have proved that,

$ABCDisaparallelogram.$