Theorem the tagent at any point …
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Sia ? 6 years, 4 months ago
GIVEN : A circle C (0, r) and a tangent AB at a point P.
TO PROVE : {tex}O P \perp A B{/tex}
CONSTRUCTION Take any point Q, other than P, on the tangent AB. Join OQ. Suppose OQ meets the circle at R.
PROOF : We know that among all line segments joining the point O to a point on AB, the shortest one is perpendicular to AB.
To prove that {tex}O P \perp A B{/tex}, it is sufficient to prove that OP is shorter than any other segment joining O to any point of AB.
Clearly, {tex}OP = OR{/tex}
Now, {tex}OQ = OR + RQ{/tex}
{tex}\Rightarrow \quad O Q > O R{/tex}
{tex}\Rightarrow \quad O Q > O P \quad [ \because O P = O R ]{/tex}
{tex}\Rightarrow \quad O P < O Q{/tex}
{tex}\therefore{/tex} {tex}O P \perp A B{/tex}
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