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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.

 

            

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Draw  a ray BL \perpPQ and CM \perp RS
Since PQ || RS (Given)
So, BL || CM and BC is a transversal
\therefore \angleLBC =  \angleMCB (Alternate interior angles).............(i)

It is known that, angle of incidence  = angle of reflection
So, \angleABL = \angleLBC and \angleMCB =  \angleMCD
\Rightarrow \angle ABL =\angle MCD..................(ii)

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

\angleABC = \angleDCB
Both the interior angles are equal
Hence AB || CD

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manish painkra

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