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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 BE ⊥ PQ and at point C, draw 

CF ⊥ RS. 

∠1 = ∠2 …(i) 

(Angle of incidence is equal to angle of reflection) 

∠3 = ∠4 …(ii) [Same reason] 

Also, ∠2 = ∠3 ... (iii) [Alternate angles] 

⇒ ∠1 = ∠4 [From (i), (ii), and (iii)] 

⇒ 2∠1 = 2∠4 

⇒ ∠1 + ∠1 = ∠4 + ∠4 

⇒ ∠1 + ∠2 = ∠3 + ∠4 [From (i) and (ii)] 

⇒ ∠BCD = ∠ABC Hence, AB || CD. [Alternate angles are equal] Proved. 

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