Question

again reflects back along CD. Prove that
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
AB ICD.
​

Answer 1

Answer:

You should first clear your question..

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