If A, B are any two points on the circle S, prove that the chord AB lies entirely within the circle.

Ashani Dasgupta Answered question

Proof: Suppose O is the center of the circle S. Join OA and OB.

OA = OB = r (radius).

P be any point on AB. It is sufficient to show OP r would imply it is outside the circle).

Clearly

- either both are right angles
- or one of these angles is obtuse.

WLOG suppose

This implies OP < OB = r hence P is inside the circle.

As P is an arbitrary point on the chord AB, this shows all points on the chord (except A and B) are inside the circle.

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