Question

it is the figures ...........
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.




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Answer 1

Answer:

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

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