Question

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

Answer 1

In the given figure,

$\angle PQD+\angle QPB={180}^{\circ }$ ......(1)

Since $PT$ and $QT$ are bisectors of $\angle BPQ$ and $\angle PQD,$

Hence,

$\frac{1}{2}\angle QPB=\angle QPT$, $\frac{1}{2}\angle DQP=\angle TQP$

Dividing equation(1) in both side by $2$

$\frac{1}{2}\times (\angle PQD+\angle QPB)={90}^{\circ }$

$(\angle PQT+\angle QPT)={90}^{\circ }$ ......(2)

Now, in $△PQT,$

$\angle PQT+\angle QPT+\angle PTQ={180}^{\circ }$

${90}^{\circ }+\angle PTQ={180}^{\circ }$ [fromequation(2)]

$\angle PTQ={90}^{\circ }$

Hence proved.