Question

in figure PQ and Rs are two Mirrors placed parallel to each other and incident ray a b strikes the mirror PQ at B , the reflected ray moves along the path BC and strikes the mirror are RS at C and again reflects back along CD prove that ab parallel to CD

Answer 1

Draw BM perpendicular to PQ and CN perpendicular to RS.

→ BM || CN

<MBC = <NCB -[1](alt. angles)

<ABM = <MBC- [2](Angle of incidence

<BCN = <NCD -[3] =Angle of

reflection)

From 1,2,3

<ABM=<NCD -[4]

1 + 4

→<MBC + <ABM = <NCD+<NCD

<ABC = <BCD

Since alt. Angles are equal

AB || CD

Answer 2

Solutions:

Draw BE and CF normals to the mirrors PQ and RS at B and C respectively.

Then, BE ⊥ PQ and CF ⊥ RS.

Since, BE and CF are perpendicular to parallel lines PQ and RS respectively. Therefore, BE || CF.

Since, BE || CF and transversal BC intersects BE and CF at B and C respectively.

Hence, ∠3 = ∠2 .............. [Alternate angles]..... (i)

But, ∠3 = ∠4 and ∠1 = ∠2 ........... [Since, angle of incidence = angle of reflection] ........ (ii)

=> ∠4 = ∠1

=> ∠3 + ∠4 = ∠2 + ∠1 .......... [Adding corresponding sides of (i) and (ii)]

=> ∠ABC = ∠BCD

Thus, transversal BC intersects lines AB and CD such that alternate interior angles ∠ABC and ∠BCD are equal. Hence, AB || CD