the length of tangent for an external point to a circle are equal .
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To prove PT=QTPT=QT
Proof: 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=QT
Hence Proved.
Proof: 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=QT
Hence Proved.
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