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Answer :
Draw perpendiculars BE and CF on PQ and RS, respectively.
Therefore, we can say that BE || CF
As angle of incidence = angle of reflection,
we get, \(\angle{a}\) = \(\angle{b}\) ....(i)
Similarly, we get, \(\angle{x}\) = \(\angle{y}\) ....(ii)
By Alternate angles theorem, \(\angle{b}\) = \(\angle{y}\)
Now, doubling the angles, we get,
2\(\angle{b}\) = 2\(\angle{y}\)
\(\Rightarrow \) \(\angle{b}\) + \(\angle{b}\) = \(\angle{y}\) + \(\angle{y}\)
from (i) and (ii),
\(\angle{a}\) + \(\angle{b}\) = \(\angle{x}\) + \(\angle{y}\)
Hence, \(\angle{ABC}\) = \(\angle{DCB}\)
Thus by converse of alternate angles theorem,
we get, AB || CD
Hence, it is proved.