Question

in figure 6.33 PQ and Rs are two Mirrors placed parallel to each other. an incident ray AB strike the mirror PQ at B, the refracted ray moves along the path BC and strikes the mirror Ras at sea and again reflects back along CD prove that AB||CD​​

Answer 1

Answer:

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Step-by-step explanation:

Let draw BM ⊥ PQ and CN ⊥ RS.

Given that  PQ || RS so that  BM || CN

Use the property of Alternate interior angles

∠2 = ∠3          … (1)

∠ABC = ∠1 +  ∠2

But ∠1 =  ∠2 so that

∠ABC = ∠2 +  ∠2

∠ABC = 2∠2

Similarly

∠BCD = ∠3 +  ∠4

But ∠3 =  ∠4 so that

∠ BCD = ∠3 +  ∠3

∠ BCD = 2∠3

From equation first

∠ABC = ∠DCB

These are alternate angles so that AB || CD

Hence proved

Answer 2

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.