Two parallel lines l and m are intersected by a transversal t. Sh…

Question

Two parallel lines l and m are intersected by a transversal t. Show that the quadrilateral formed by the bisectors of interior angles is a rectangle.

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Answer

l || m and t is a transversal.$\Rightarrow\angle\text{APR}=\angle\text{PRD}$ (alternate angles)
$\Rightarrow\frac{1}{2}\angle\text{APR}=\frac{1}{2}\angle\text{PRD}$
$\Rightarrow\angle\text{SPR}=\angle\text{PRQ}$ (PS and RQ are the bisectors of $\angle\text{APR}$ and $\angle\text{PRD}$)
Thus, PR intersects PS and RQ at P and R respectively such that $\angle\text{SPR}=\angle\text{PRQ}$ i.e., alternate angles are equal.$\Rightarrow\text{PS || RQ}$
Similarly, we have $\text{SR ∥ PQ.}$ Hence, PQRS is a parallelogram. Now, $\angle\text{BPR}+\angle\text{PRD}=180^{\circ}$ (interior angles are supplementary)$\Rightarrow2\angle\text{QPR}+2\angle\text{QRP}=180^{\circ}$ (PQ and RQ are the bisectors of $\angle\text{BPR}$ and $\angle\text{PRD}$)
$\Rightarrow\angle\text{QPR}+\angle\text{QRP}=90^{\circ}$
In $\triangle\text{PQR},$ by angle sum property,$\angle\text{PQR}+\angle\text{QPR}+\angle\text{QRP}=180^{\circ}$
$\Rightarrow\angle\text{PQR}+90^{\circ}=180^{\circ}$
$\Rightarrow\angle\text{PQR}=90^{\circ}$
Since PQRS is a parallelogram,$\angle\text{PQR}=\angle\text{PSR}$
$\Rightarrow\angle\text{PSR}=90^{\circ}$
Now, $\angle\text{SPQ}+\angle\text{PQR}=180^{\circ}$ (adjacent angles in a parallelogram are supplementary)$\Rightarrow\angle\text{SPQ}+90^{\circ}=180^{\circ}$
$\Rightarrow\angle\text{SPQ}=90^{\circ}$
$\Rightarrow\angle\text{SRQ}=90^{\circ}$
Thus, all the interior angles of quadrilateral PQRS are right angles. Hence, PQRS is a rectangle.