Question

in figure PQ and Rs are two Mirrors placed parallel to each other and incident ray be strikes the mirror PQ and be reflected ray moves along the path DC and strikes the mirror are SSC and again reflects back around CD prove that a be parallel to CD

Answer 1

heya...


here is the ans..


given that- pq // rs

to prove- ab // cd

construction- draw bm and bn normal to the mirror

proof- <abn = < nbz ( law of reflection)- (1)

similarly < bcn = < ncd ..

and we know that bm // cn - perpendicular drawn on parallel lines are parallel

therefore < nbc = < bcn (alternate interior

.... angles) - (2)



there from (1) and (2)


< abn = < ncd

now add <bnc = ncd both side..

we will get ..< abc = < bcd


alternate angles are equal...
therefore 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