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Prove that out of all chords which passesthrough any point circle, that chord will besmallest which is perpendicular on diameterwhich passes through that point. |
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Answer» Let C(O, r) is a circle and let M is a point within it where O is the centre and r is the radius of the circle. Let CD is another chord passes through point M. We have to prove that AB < CD. Now join OM and draw OL perpendicular to CD. In right angle triangle OLM, OM is the hypotenuse. So OM > OL => chord CD is nearer to O in comparison to AB. => CD > AB => AB < CD So all chords of a circle of a circle at a given point within it, the smallest is one which is bisected at that point. |
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