How to prove that any point on the perpendicular bisector of a line segment AB is equidistant from points A and B.
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Let the perpendicular bisector be CD with the point of intersection of AB and CD being O. So we have AO = OB (CD is a bisector of AB).
From point C draw CB and CA. Consider triangles CAO and CBO.
SAS congruence exists between them, as CO is common, angle COA = angle COB = 90 deg. and AO = OB.
Hence, CA = CB.
For any point C this is true.
From point C draw CB and CA. Consider triangles CAO and CBO.
SAS congruence exists between them, as CO is common, angle COA = angle COB = 90 deg. and AO = OB.
Hence, CA = CB.
For any point C this is true.
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