In figure, `oversetfrown(AB)congoversetfrown(AC)` and O is the centre of the circle,Prove that OA is the perpendicualr bisector of BC.

Explanation

Since `oversetfrown(AB)congoversetfrown(AC)` (given)
`rArrAB=AC` (corresponding chords of congruent arcs are equal)
In `DeltaOAB` and `DeltaOAC`,
`because" "{(AB,=,,AC,("just proved")),(OA,=,,OA,("common")),(OB,=,,OC,("each radii")):}`
`thereforeDeltaOABcongDeltaOAC` (by SSS congruency)
`therefore angle1 =angle2` (c.p.c.t)
Now, in `Delta OMB` and `DeltaOMC`,
`because" "{(OB,=,,AC,("each radii")),(angle1,=,,angle2,("just proved")),(OM,=,,OM,("common")):}`
`therefore Delta OMBcongDeltaOMC` (by S.A.S. congruency)
`therefore BM=MC` (c.p.c.t) .......(1)
Also, `angle3=angle4` (c.p.c.t)
But `angle3+angle4=180^@` (L.P.A.)
`rArrangle3=angle4=90^@` ......(2)
From (1) and (2) , we get
OM is the perpendicular bisector of BC.
`rArrOA` is the perpendicular bisector of BC.