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Given 2 2circles intersect at a and b T.P the line formed by their centers(o and p) is the perpendicular bisector of chord ab PROOF con.tri. oap and obp oa=ob (radii of same circle) pb=pa ap=pa (common) therefore, oap is cong. to obp by SSS let m the point of intersection of ab and op join om oa=ob aom=bom (proved by CPCT) om=om (common) therefore tri. aom is cong. to bom by sas hence, am=bm amo=bmo now ang. amo+bmo=180 (linear pair) as amo=bmo, 2 amo=180 amo=90 hence, bmo=90 lly we can prove it for the other two. i.e all angles formed r 90 HENCE THE PROOF..

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See diagram.

AB is the line joining intersections of the circles with centers O and P. Let OC be the perpendicular onto AB from O.

OA = OB as OA and OB are radii of circle with center O. angle OCA = angle OCB = 90 deg. OC is a common side. Hence two triangles OCA and OCB are congruent. Hence AC = CB. So O lies on perpendicular bisector of AB.

PA = PB as they are radii of circle with center P. draw a perpendicular from P onto AB. Let it meet at C'. Now between the two triangles AC'P and BC'P, PC' is common side, angle AC'P = angle BC'P = 90 deg. Hence the two triangles are congruent. Hence AC' = BC'. SO P lies on the perpendicular bisector of AB.