Question

6. In Fig. 6.33, PO 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 Cand
again reflects back along CD. Prove that
ABCD.​

Answer 1

Answer:

it answer is in the the picture attached

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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