Question

Prove that every point on the perpendicular bisector of AB is equidistant from A and B.

Answer 1

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.

therefore ,For any point C this is true.

Answer 2

Answer:

: P is equidistant from A and B. ∆BOP (By S.A.S.) => AP = BP ( By C.P.C.T. ) Hence , "P" is equidistant from A and B