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in Plane Geometry and Line and Angle by (35.3k points)
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In the 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 and again reflects back along CD. Prove that AB || CD.

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At point B, draw BM ⊥ RS and at point C, draw CN ⊥ PQ

∠1 = ∠2 …(i)

and ∠3 = ∠4 ,..(ii)

(angle of incidence is equal to angle of reflection)

Also ∠2 = ∠3 …(iii)

(Alternate angles)

From (i), (ii) and (iii), we get

⇒ ∠1 = ∠4

⇒ 2∠1 = 2∠4

⇒ ∠1 + ∠1 = ∠4 + ∠4

⇒ ∠1 + ∠2 = ∠3 + ∠4

[using (i) and (ii)]

⇒ ∠ABC = ∠BCD

(Alternate angles)

⇒ AB || CD

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