Provd that the lengths of the tangents from am external points toa circle are equal.
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To prove PT=QTPT=QTProof: Consider the triangle OPTOPT and OQTOQT.OP=OQOP=OQ∠OPT=∠OQT=90∘∠OPT=∠OQT=90∘OT=OTOT=OT (common side)Hence by RHS the triangles are equal.Hence PT=QTPT=QTHence Proved.
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