Answers

2015-02-22T12:49:16+05:30
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..



2 3 2
2015-02-23T01:05:58+05:30

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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.

1 5 1