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Question

Show that the tangents at the extremities of a chords of a circle makes equal angles with the chord.


Solution


Let $$PQ$$ be the chord of a circle with center $$O$$.
Let $$AP$$ and $$AQ$$ be the tangents at points $$P$$ and $$Q$$ respectively.

Let us assume that both the tangents meet at point $$A$$.
Join points $$O,P$$. let $$OA$$ meets $$PQ$$ at $$R$$
Here we have to prove that $$\angle APR = \angle AQR$$

Consider, $$\Delta APR$$ and $$\Delta AQR$$
$$AP=AQ$$ (Tangents drawn from an internal point to a circle are equal0
$$\angle PAR = \angle QAR$$
$$AR=AR$$ {common side}
$$\therefore \Delta APR \cong \Delta AQR$$ [SAS congruence criterion]

Hence,
$$\angle APR = \angle AQR\left[ {CPCT} \right]$$

1219024_1308822_ans_3f9e65319f8f46e4aa0b8cbdc17abf8e.PNG

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