If A, B are any two points on the circle S, prove that the chord AB lies entirely within the circle.
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).
- either both are right angles
- or one of these angles is obtuse.
WLOG suppose . Clearly in it is the largest angle. Hence side opposite to is the largest side in . Therefore OB > OP.
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.