Question

Show that if there is a transversal between two parallel lines, the interior angle bisectors meet to form a rectangle. [4 MARKS]

Answer 1

Diagram : 1 Mark
Concept : 1 Mark
Proof : 2 Marks

Consider 2 parallel lines $l$ and $m$.

It is given that $PS\parallel QR$ and a transversal intersect them at points $A$ and $C$ respectively.

The bisector of $\angle PAC$ and $\angle ACQ$ intersect at $B$ and bisectors of $\angle ACR$ and $\angle SAC$ intersect at $D.$

We are to show that quadrilateral ABCD is a rectangle.

Now, $\angle PAC=\angle ACR$ [Pair of alternate angles]

$\Rightarrow \frac{1}{2}\angle PAC=\frac{1}{2}\angle ACR$

i.e. $\angle BAC=\angle ACD$

These form a pair of alternate angles for lines $AB$ and $DC$ with $AC$ as transversal and they are equal also.

$\Rightarrow AB||DC$ .......(i)

Similarly, BC|| AD.......(ii) [Considering $\angle ACB$ and $\angle CAD$]

$\therefore$ quadrilateral ABCD is a parallelogram.

Also, $\angle PAC+\angle CAS={180}^{\circ }$ [linear pair]

$\Rightarrow \angle BAC+\angle CAD={90}^{\circ }$

$\Rightarrow \angle BAD={90}^{\circ }$

So, ABCD is a parallelogram in which one angle is ${90}^{\circ }$.

$\therefore$ ABCD is a rectangle.