Question

Prove that if two parallel lines are intersected by a transversal, then prove that the bisectors of the interior angles form a rectangle.

Answer 1

Given: Two parallel lines AB and CD and a transversal EF intersect them at G and H respectively. GM, HM, GL and HL are the bisectors of the two pairs of interior angles.
To Prove: GMHL is a rectangle.
Proof:

$\because AB\parallel CD$

$\therefore \angle AGH=\angle DHG$ (Alternate interior angles)

$\Rightarrow \frac{1}{2}\angle AGH=\frac{1}{2}\angle DHG$

$\Rightarrow \angle 1=\angle 2$
(GM & HL are bisectors of $\angle AGH$ and $\angle DHG$ respectively)

$\Rightarrow GM\parallel HL$
($\angle 1$ and $\angle 2$ from a pair of alternate interior angles and are equal)

Similarly, $GL\parallel MH$

So, GMHL is a parallelogram.

$\because AB\parallel CD$

$\therefore \angle BGH+\angle DHG={180}^o$

(Sum of interior angles on the same side of the transversal $={180}^o$)

$\Rightarrow \frac{1}{2}\angle BGH+\frac{1}{2}\angle DHG={90}^o$

$\Rightarrow \angle 3+\angle 2={90}^o$ .....(3)

(GL & HL are bisectors of $\angle BGH$ and $\angle DHG$ respectively).

In $\Delta GLH,\angle 2+\angle 3+\angle L={180}^o$

$\Rightarrow {90}^o+\angle L={180}^o$ Using (3)

$\Rightarrow \angle L={180}^o-{90}^o$

$\Rightarrow \angle L={90}^o$

Thus, in parallelogram GMHL, $\angle L={90}^o$

Hence, GMHL is a rectangle.