Question

6. In Fig. 6.33, PQ 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 C and
again reflects back along CD. Prove that
AB|CD
mp​

Answer 1

Answer:

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