Question

in the following figure PQ and RS are two Mirrors placed parallel to each other and incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirrorRSat C and again reflect back alongCD. prove that AB||CD

Answer 1

its so simple ..

according to physics,, angle of incidence is always equal to angle of reflection..

so ,, here in the figure
AB is the incident ray .. and CD is the reflected ray

so ,, AB=CD( proved )

plz mark It as brainliest

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