Question

In the given Figure, line AB || line CD and line PQ is the transversal. Ray PT and ray QT are bisectors of $\angle BPQ\text{and}\angle PQD\text{respectively.}$
$\text{Prove that m}\angle PTQ={90}^{\circ }.$

Answer 1

$AB||CD\text{and}PQ$ is a transversal line.
$\angle BPQ+\angle DQP={180}^{\circ }$ (Angles on the same side of a transversal line are supplementary angles)
$\Rightarrow \frac{1}{2}\angle BPQ+\frac{1}{2}\angle DQP=\frac{180}{2}$
$\Rightarrow \angle QPT+\angle PQT={90}^{\circ }$
$\text{In}△PQT$
$\angle QPT+\angle PQT+\angle PTQ={180}^{\circ }\text{(Angle sum property)}$
$\Rightarrow {90}^{\circ }+\angle PTQ={180}^{\circ }$
$\Rightarrow \angle PTQ={90}^{\circ }$
Hence proved.