Question

Let $\overleftrightarrow{AB}$ and $\overleftrightarrow{CD}$ be two parallel lines and $PQ$ be a transversal. Show that the angle bisectors of a pair of two internal angles on the same side of the transversal are perpendicular to each other.

Answer 1

It is given $AB||CD$

$PQ$ is transversal, cut $AB$ and $CD$ at $M$ and $N$ respectively.

$MX$ and $NY$ are angular bisector of $\angle BMD$ and $\angle MND$ respectively.

$\therefore \angle MON={90}^o$

From the figure,

$\angle BMN+\angle <MND={180}^o$ (pre-position)

Let $\angle BMD=2a$ and $\angle MND=2b$

$\Rightarrow 20+2b={180}^o$

$\Rightarrow a+b={90}^o$

From $△MNO$,

$\angle LMN+\angle ONM+\angle MON={180}^o$ (angle sum property)

$a+b+\angle MON={180}^o$

$\Rightarrow 90+\angle MON={180}^o$

$\therefore \angle MON={90}^o$

$\therefore$ angle bisector of a pair of two internal triangles on the same sides of the transversal are perpendicular to each other.