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. $\angle ABE=\angle FCD......(ii)$
$\angle EBC+\angle ABE=\angle BCF+\angle FCD$
$\angle ABC=\angle BCD$
Now if we take line $AB$ and $CD$ in consideration, alternate interior angle that are $\angle ABC$ and $\angle BCD$ are equal.
Therefore, $AB||CD$ You're just one step away

Subjective Medium Published on 09th 09, 2020
Questions 120418
Subjects 10
Chapters 88
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