Prove that the tangent drawn at …
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Sia ? 4 years, 10 months ago
Given: AB is arc in circle C(0, r), P is mid-point of arc AB; XY is tangent to the circle at P.
To Prove: AB II XY. Join OA and OB.
Here {tex}\angle{/tex}AOP = {tex}\angle{/tex}BOP(angle subtended by equal arc)
{tex}OA = OB, OC = OC{/tex}
{tex}\triangle{/tex}ACO {tex}\cong{/tex} {tex}\triangle{/tex}BCO (SAS)
{tex}\Rightarrow{/tex} {tex}AC = BC {/tex}
{tex}\Rightarrow{/tex} {tex}\mathrm { OC } \perp \mathrm { AB }{/tex}
(line joining mid-point of chord with centre of the circle is perpendicular to the chord)Also {tex}\angle \mathrm { OPY } = 90 ^ { \circ }{/tex}
{tex}\Rightarrow \angle \mathrm { OCB } = \angle \mathrm { OPY }{/tex} (corresponding angles)
{tex}\therefore{/tex} {tex}AB\ || \ XY{/tex}
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