Question

In Fig. 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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Answer 1

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Draw BL⊥PQ and CM⊥RS. As PQ∥RS, so, BL∥CM.

The alternate interior angles are equal, so,

∠LBC=∠MCB (1)

It is known that the angle of reflection is equal to the angle of incidence, therefore,

∠ABL=∠LBC (2)

And,

∠MCB=∠MCD (3)

From equation (1), (2) and (3),

∠ABL=∠MCD (4)

Add equation (1) and (4),

∠LBC+∠ABL=∠MCB+∠MCD

∠ABC=∠BCD

Since, these are the interior angles and are equal, hence, AB∥CD.

Answer 2

Draw BL⊥PQ and CM⊥RS. As PQ∥RS, so, BL∥CM.

The alternate interior angles are equal, so,

∠LBC=∠MCB (1)

It is known that the angle of reflection is equal to the angle of incidence, therefore,

∠ABL=∠LBC (2)

And,

∠MCB=∠MCD (3)

From equation (1), (2) and (3),

∠ABL=∠MCD (4)

Add equation (1) and (4),

∠LBC+∠ABL=∠MCB+∠MCD

∠ABC=∠BCD

Since, these are the interior angles and are equal, hence, AB∥CD.