Question

In the given figure, 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

Step-by-step explanation:

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▪️Since BE and FC are normal to PQ and RS respectively, therefore, BE ∣∣ FC

▪️Let, ∠ABE=∠EBC= x ( PQ is a mirror so angle of incidence is equal to angle of reflection)

▪️∠FCD =∠BCF=y (RS is a mirror so angle of incidence is equal to angle of reflection)

Now considering BE and FC, taking BC as transversal,

▪️∠EBC=∠BCF...(i) (alternate interior angle)

▪️i.e. x=y

▪️i.e. ∠ABE=∠FCD....(ii)

▪️adding equation (i) and (ii)

▪️∠EBC+∠ABE =∠BCF+∠FCD

▪️∠ABC =∠BCD

▪️Now if we take line AB and CD in consideration, alternate interior ∠ that are ∠ABC and ∠BCD are equal.

Therefore, AB ∣∣ CD

Answer 2

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Let us draw lines BX and CY that are parallel to each other, to get AB || CD.

Let, ∠ABE=∠EBC= x ( PQ is a mirror so angle of incidence is equal to angle of reflection)

➞ ️∠FCD =∠BCF=y (RS is a mirror so angle of incidence is equal to angle of reflection)

Now considering BE and FC, taking BC as transversal,

➞∠EBC=∠BCF...(i) (alternate interior angle)

➞ i.e. ∠ABE=∠FCD....(ii)

➞ adding equation (i) and (ii)

➞ ∠EBC+∠ABE =∠BCF+∠FCD

➞ ∠ABC =∠BCD

➞ ️Now if we take line AB and CD in consideration, alternate interior ∠ that are ∠ABC and ∠BCD are equal.

Therefore, AB || CD