Prove that the chord which passes through a point inside the circle is the smallest amongst all the chords and is perpendicular to the diameter passing through that point.
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Given : AB is diameter passing through point R inside circle of center O and required chord is PQ which is perpendicular to AB,CD is other chord passing through R. To Prove : Chord PQ is the smallest. Construction : Through O, draw OM⊥CD Proof : In right angled ΔOMR Hypotenuse OR is longest side of ΔOMR OR>OM OR is distance of chord PQ from center O and OM is distance of chord CD from center O. Since any one of two chords of a circle, the larger chord is nearer to the center. Thus, chord PQ is the smallest