Mathematics

In the adjacent figure PQ and RS are two mirrors placed parallel to each other. An incident ray $$\overrightarrow{AB}$$ strikes the mirror PQ at B, the reflected ray moves along the path $$\overrightarrow{BC}$$ and strikes the mirror RS at C and again reflected back along $$\overrightarrow{CD}$$. Prove that AB || CD.


SOLUTION
Since $$BE$$ and $$FC$$ are normal to $$PQ$$ and $$RS$$ respectively, therefore, $$BE||FC$$
Let, $$\angle ABE=\angle EBC=x(PQ$$ is a mirror so angle of incidence is equal to angle of reflection)
$$\angle FCD=\angle BCF=y$$ (RS is a mirror so angle of incidence is equal to angle of reflection)
Now considering BE and FC, taking BC as transversal,
$$\angle EBC=\angle BCF........(i)$$ (alternate interior angle)
i.e. $$x=y$$
i.e. $$\angle ABE=\angle FCD......(ii)$$
adding equation (i) and (ii)
$$\angle EBC+\angle ABE=\angle BCF+\angle FCD$$
$$\angle ABC=\angle BCD$$
Now if we take line $$AB$$ and $$CD$$ in consideration, alternate interior angle that are $$\angle ABC$$ and $$\angle BCD$$ are equal.
Therefore, $$AB||CD$$
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