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 || CD.​

Answer 1

Step-by-step explanation:

Let draw BM ⊥ PQ and CN ⊥ RS.

Given that  PQ || RS so that  BM || CN

Use the property of Alternate interior angles

∠2 = ∠3          … (1)

∠ABC = ∠1 +  ∠2

But ∠1 =  ∠2 so that

∠ABC = ∠2 +  ∠2

∠ABC = 2∠2

Similarly

∠BCD = ∠3 +  ∠4

But ∠3 =  ∠4 so that

∠ BCD = ∠3 +  ∠3

∠ BCD = 2∠3

From equation first

∠ABC = ∠DCB

These are alternate angles so that AB || CD

Hence proved