In fig 6.33 PQ and RS are two mirrors placed parallel to each other.Anincident ray AB strike the mirror PQ at B,the reflected ray moves along a path BC and striks the mirror RS at C and again reflects back long CD. prove that AB // CD.
Answers
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
Answer:
Step-by-step explanation:
PQ || RS ⇒ BL || CM
[∵ BL || PQ and CM || RS]
Now, BL || CM and BC is a transversal.
∴ ∠LBC = ∠MCB …(1) [Alternate interior angles]
Since, angle of incidence = Angle of reflection
∠ABL = ∠LBC and ∠MCB = ∠MCD
⇒ ∠ABL = ∠MCD …(2) [By (1)]
Adding (1) and (2), we get
∠LBC + ∠ABL = ∠MCB + ∠MCD
⇒ ∠ABC = ∠BCD
i. e., a pair of alternate interior angles are equal.
∴ AB || CD.