Fix a real number l and two rays in the plane with common endpoint S. Point A moves along one ray, while point B moves along the other ray, such that AS+SB=l. Prove that the perpendicular bisector of AB passes through a fixed point in the plane.
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Let P, Q be the points on rays SA, SB respectively such that SP=l=SQ. I claim the point is Circumcenter of \triangle SPQ.
proof: Let O be that circumcenter. \angle OSB = \angle OSQ = \angle OPS = \angle OPA, OP = OQ = OS, and finally SB = l-AS = l-(l-AP) = AP. Now by SAS, we find that \triangle OSB \equiv \triangle OPA and thus OA = OB. this is equivalent to the fact that O lies on the perpendicular bisector of AB.
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