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 that AB || CD.
Answers & Comments
Step-by-step explanation:
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