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

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