Answer:
Given, a circle with centre O from point P with tangents PQ and PR drawn. To prove: \[PQ=PR\] Proof: In \[\Delta \text{ }PQO\] and \[\Delta \text{ }PRO\] \[\angle PQO=\angle PRO=90\] (Angle between radius and tangent) \[OQ=OR\] (Radius) \[OP\] is common \[\therefore \] By R.H.S. congruence rule \[\Delta \,PQO\cong \Delta \,PRO\] \[PQ=PR\] (c.p.c.t.) Hence Proved.
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