prove that diameter is the longest chord in the circle

Let AB be the diameter of a circle C (0, r) and let CD be any other chord.

AB is nearer to the centre than CD.

AB > CD  (Of any two chords of a circle, the one which is nearer to the centre is larger.)

Hence, AB is larger than every other chord of the circle.

 

 

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Take any chord in a circle, say with endpoints AB. Let O be the center of the circle. Then segments AO and BO are radii of the circle. AOB is a triangle, and we know that the sum of the lengths of two sides of a triangle is always greater than or equal to the length of the third side. So:

|AB| ≤ |AO| + |BO|

(Here |AB| means length of AB.) Now the length of a diameter is twice the length of a radius, which is precisely equal to the quantity |AO| + |BO|.

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