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

Given: l || m and the bisectors of interior angles intersect at B and D.

To prove: ABCD is a rectangle.

Proof:

Since, l || m (Given)

So, $\angle PAC=\angle ACR$ (Alternate interior angles)

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

$\Rightarrow \angle BAC=\angle ACD$

but, these are a pair of alternate interior angles for AB and DC.

$\Rightarrow AB\parallel DC$

Similarly, $BC\parallel AD$

So, ABCD is a parallelogram.

Also,
$\angle PAC+\angle CAS=180°$ (Linear pair)

$$\begin{gathered} \Rightarrow \frac{1}{2}\angle PAC+\frac{1}{2}\angle CAS=90° \\ \Rightarrow \angle BAC+\angle CAD=90° \\ \Rightarrow \angle BAD=90° \end{gathered}$$

But, this an angle of the parallleogram ABCD.

Hence, ABCD is a rectangle.