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 1

Let $PQ,SP,RQ$ and $SR$ be the angle bisectors of interior angles as shown in the figure. Let the quadrilateral so formed be $PQRS$.

Now,

Since, $PQ$ is the angle bisector of $\angle BPR$, therefore,

$\Rightarrow \angle BPQ=\angle QPR$ ..(i)

Similarly, we can say that,

$\Rightarrow \angle APS=\angle SPR$ ..(ii)

$\Rightarrow \angle DRQ=\angle QRP$ ..(iii)

$\Rightarrow \angle CSR=\angle SRP$ ..(iv)

Also, $\angle APR=\angle DRP$ [Pair of alternate interior angles]

$\Rightarrow 2\angle APS=2\angle QRP$ [From (ii) and (iii)]

$\Rightarrow \angle APS=\angle QRP$

But these two angles also form a pair of alternate interior angles.

Therefore, $PS\parallel QR$..(v)

Similarly we can say that $SR\parallel QP$..(vi)

From (v) and (vi), $PQRS$ is a parallelogram.

Also,$\angle BPR+\angle DRP={180}^{\circ }$ [Angle on the same side of the transversal are supplementary]

$\Rightarrow 2\angle QPR+2\angle QRP={180}^{\circ }$ [From (i) and (iii)]

$\Rightarrow \angle QPR+\angle QRP={90}^{\circ }$ ...(vii)

Now, in $\Delta PQR$

$\Rightarrow \angle QPR+\angle QRP+\angle PQR={180}^{\circ }$

$\Rightarrow \angle PQR={90}^{\circ }$ [From (vii)]

Since one of the angle of the parallelogram $PQRS$ is ${90}^{\circ }$, therefore, $PQRS$ is a rectangle.