# 6.  In Fig. 6.33, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB $||$ CD.

Draw  a ray BL $\perp$PQ and CM $\perp$ RS
Since PQ || RS (Given)
So, BL || CM and BC is a transversal
$\therefore$ $\angle$LBC =  $\angle$MCB (Alternate interior angles).............(i)

It is known that, angle of incidence  = angle of reflection
So, $\angle$ABL = $\angle$LBC and $\angle$MCB =  $\angle$MCD
$\Rightarrow \angle ABL =\angle MCD$..................(ii)

Adding eq (i) and eq (ii), we get

$\angle$ABC = $\angle$DCB
Both the interior angles are equal
Hence AB || CD

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