Question

14.
Do
PE
B
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
A
along the path BC and strikes the mirror RS at C
and again reflected back along CD. Prove that
AB || CD.
RE
C​

Answer 1

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.