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Quadrilaterals — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsQuadrilaterals5 Marks
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.
✓

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.

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