Question

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 II CD.

Answer 1

Let us draw BM ⊥ PQ and CN ⊥ RS.

Let us draw BM ⊥ PQ and CN ⊥ RS.As PQ || RS,

Let us draw BM ⊥ PQ and CN ⊥ RS.As PQ || RS,Therefore, BM || CN

Let us draw BM ⊥ PQ and CN ⊥ RS.As PQ || RS,Therefore, BM || CNThus, BM and CN are two parallel lines and a transversal line BC cuts them at B and C respectively.

Let us draw BM ⊥ PQ and CN ⊥ RS.As PQ || RS,Therefore, BM || CNThus, BM and CN are two parallel lines and a transversal line BC cuts them at B and C respectively.∴∠2 = ∠3 (Alternate interior angles)

Let us draw BM ⊥ PQ and CN ⊥ RS.As PQ || RS,Therefore, BM || CNThus, BM and CN are two parallel lines and a transversal line BC cuts them at B and C respectively.∴∠2 = ∠3 (Alternate interior angles)However, ∠1 = ∠2 and ∠3 = ∠4 (By laws of reflection)

Let us draw BM ⊥ PQ and CN ⊥ RS.As PQ || RS,Therefore, BM || CNThus, BM and CN are two parallel lines and a transversal line BC cuts them at B and C respectively.∴∠2 = ∠3 (Alternate interior angles)However, ∠1 = ∠2 and ∠3 = ∠4 (By laws of reflection)∴ ∠1 = ∠2 = ∠3 = ∠4

Let us draw BM ⊥ PQ and CN ⊥ RS.As PQ || RS,Therefore, BM || CNThus, BM and CN are two parallel lines and a transversal line BC cuts them at B and C respectively.∴∠2 = ∠3 (Alternate interior angles)However, ∠1 = ∠2 and ∠3 = ∠4 (By laws of reflection)∴ ∠1 = ∠2 = ∠3 = ∠4Also, ∠1 + ∠2 = ∠3 + ∠4

Let us draw BM ⊥ PQ and CN ⊥ RS.As PQ || RS,Therefore, BM || CNThus, BM and CN are two parallel lines and a transversal line BC cuts them at B and C respectively.∴∠2 = ∠3 (Alternate interior angles)However, ∠1 = ∠2 and ∠3 = ∠4 (By laws of reflection)∴ ∠1 = ∠2 = ∠3 = ∠4Also, ∠1 + ∠2 = ∠3 + ∠4∠ABC = ∠DCB

Let us draw BM ⊥ PQ and CN ⊥ RS.As PQ || RS,Therefore, BM || CNThus, BM and CN are two parallel lines and a transversal line BC cuts them at B and C respectively.∴∠2 = ∠3 (Alternate interior angles)However, ∠1 = ∠2 and ∠3 = ∠4 (By laws of reflection)∴ ∠1 = ∠2 = ∠3 = ∠4Also, ∠1 + ∠2 = ∠3 + ∠4∠ABC = ∠DCBHowever, these are alternate interior angles.

Let us draw BM ⊥ PQ and CN ⊥ RS.As PQ || RS,Therefore, BM || CNThus, BM and CN are two parallel lines and a transversal line BC cuts them at B and C respectively.∴∠2 = ∠3 (Alternate interior angles)However, ∠1 = ∠2 and ∠3 = ∠4 (By laws of reflection)∴ ∠1 = ∠2 = ∠3 = ∠4Also, ∠1 + ∠2 = ∠3 + ∠4∠ABC = ∠DCBHowever, these are alternate interior angles.∴ AB || CD

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Answer 2

Step-by-step explanation:

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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.