Question
In 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.
Solution
Construction: Draw ray BL ⊥ PQ and ray CM ⊥ RS.
Proof: ∵ BL ⊥ PQ, CM ⊥ RS and PQ || RS ∴ BL || CM
∠LBC = ∠MCB ...(1)
| Alternate Interior Angles
∠ABL = ∠LBC ...(2)
| ∵ Angle of incidence = Angle of reflection
∠MCB = ∠MCD ...(3)
| ∵ Angle of incidence = Angle of reflection
From (1), (2) and (3), we get
∠ABL = ∠MCD ...(4)
Adding (1) and (4), we get
∠LBC + ∠ABL = ∠MCB + ∠MCD
⇒ ∠ABC = ∠BCD
But these are alternate interior angles and they are equal.
So, AB || CD.
