Question

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 II CD.
​

Answer 1

The interior angles and are equal, hence, AB∥CD.

Step-by-step explanation:

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

Step-by-step explanation:

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