Question

In a parallelogram $ABCD,$ prove that sum of any two consecutive angles is ${180}^{\circ }$

Answer 1

Since $ABCD$ is a parallelogram, $AD\parallel BC$

Now,$AD\parallel BC$ and transversal $AB$ intersects them at $A$ and $B$ respectively.

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

$\because$ Sum of the interior angles on the same side of the transversal is ${180}^{\circ }$

Since $ABCD$ is a parallelogram, $AB\parallel DC$

Now,$AB\parallel DC$ and transversal $AC$ intersects them at $A$ and $C$ respectively.

$\therefore \angle C+\angle D={180}^{\circ }$

$\because$ Sum of the interior angles on the same side of the transversal is ${180}^{\circ }$

Now,$AD\parallel BC$ and transversal $AC$ intersects them at $A$ and $C$ respectively.

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

$\because$ Sum of the interior angles on the same side of the transversal is ${180}^{\circ }$

Now,$DC\parallel AB$ and transversal $BD$ intersects them at $B$ and $D$ respectively.

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

$\because$ Sum of the interior angles on the same side of the transversal is ${180}^{\circ }$

Hence proved.