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
ABllCD.​

Answer 1

Answer: Yes AB//CD

Step-by-step explanation:

Angle of incidence of AB on mirror PQ is equals to the angle of reflection of BC to mirror RS, also the angle of incidence of BC on mirror RS is equals to the angle of reflection of CD.

Answer 2

Answer:

Step-by-step explanation:

PQ || RS ⇒ BL || CM

[∵ BL || PQ and CM || RS]

Now, BL || CM and BC is a transversal.

∴ ∠LBC = ∠MCB …(1) [Alternate interior angles]

Since, angle of incidence = Angle of reflection

∠ABL = ∠LBC and ∠MCB = ∠MCD

⇒ ∠ABL = ∠MCD …(2) [By (1)]

Adding (1) and (2), we get

∠LBC + ∠ABL = ∠MCB + ∠MCD

⇒ ∠ABC = ∠BCD

i. e., a pair of alternate interior angles are equal.

∴ AB || CD.