Question

14.
Pe
In the adjacent figure 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
RG
C С
and again reflected back along CD. Prove that
AB || CD
[Hint: Perpendiculars drawn to parallel lines are also parallel.]
​

Answer 1

Answer:

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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.