Question

In Hg 0.33, PO and RS are two mirrors placed
parallel to each other An incident ray AB strikes
the mirror 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 Il CD,​

Answer 1

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:

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.