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If a point is equidistant from the endpoints of a line segment, then it lies on the perpendicular bisector of the segment.
Based on the characteristics of the diagram, the following relation holds true.
AC=CB ⇓ CM⊥ AB and AM=MB
To prove that C lies on the perpendicular bisector of AB, it will be shown that the line perpendicular to AB through C bisects AB. If M is the point of intersection between the line and the segment, it must be proven that AM=MB.
This line forms two right triangles that share a common leg CM. Because all right angles are congruent, ∠ AMC is congruent to ∠ BMC. Also, by the Reflexive Property of Congruence, CM is congruent to itself. Since AC is equal to BC, AC is congruent to BC. ∠ AMC ≅ ∠ BMC CM≅CM AC≅BC By the Hypotenuse Leg Theorem, △ AMC and △ BMC are congruent triangles. Because corresponding parts of congruent figures are congruent, AM is congruent to BM.
Additionally, it was already known that CM and AB are perpendicular.
CM⊥ AB and AM=MB
By the definition of a perpendicular bisector, CM is the perpendicular bisector of AB. Therefore, C lies on the perpendicular bisector of AB.