Question

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 II CD. In Fig. 6.33.

Answer 1

according to law of reflection
angle of incidence = angle of reflection
construction : draw normals at b and c
let it be wb & zc
therefore ,
angle 1 = angle 2 & angle 3 = angle 4
we know that perpendicular on Parallel lines are also parallel to each other
=> WB || ZC
=>angle2 = angle3 (by alt. angle)
2angle2 = 2angle3
angleAOC = angleBCD
but these angles are alternate so
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