Question

In the adjacent 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 reflected back along CD Prove thatAB || CD. [Hint : Perpendiculars drawn to parallel lines are also parallel.]​

Answer 1

Answer:

Step-by-step explanation:

ANSWER

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 angle that are ∠ABC and ∠BCD are equal.

Therefore, AB∣∣CD

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.