Question

bo
B
PS
14.
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
A
C
along the path BC and strikes the mirror RS at C
RA
and again reflected back along CD. Prove that
AB || CD.
[Hint: Perpendiculars drawn to parallel lines are also parallel.]
2.​

Answer 1

Answer:

AB||CD(hence proved)

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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.