Question

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 ABIICD.

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Step-by-step explanation:

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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.