Question

in the figure PQ and RS are two mirror 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.​

Answer 1

Step-by-step explanation:

${\huge{\boxed{\sf{\pink{☞Answe࿐★}}}}}$

PQ || RS ⇒ BL || CM

[∵ BL || PQ and CM || RS]

Now, BL || CM and BC is a transversal.

∴ ∠LBC = ∠MCB [Alternate interior angles]

Since, angle of incidence = Angle of reflection

∠ABL = ∠LBC and ∠MCB = ∠MCD

⇒ ∠ABL = ∠MCD

Adding (1) and (2), we get

∠LBC + ∠ABL = ∠MCB + ∠MCD

⇒ ∠ABC = ∠BCD

i. e., a pair of alternate interior angles are equal.

∴ AB || CD.

$\displaystyle{\implies\underline{\boxed{\red{\sf\:ab \: ll \: cd}}}}$