# If two circles in Let two circles O and O' intersect at two points A and B so that AB is the common chord of two circles.

OO' is the line segment joining the centers.
Let OO' intersect AB at M
Now Draw line segments OA, OB , O'A and O'B
In ΔOAO' and OBO' , we have
OA = OB (radii of same circle)
O'A = O'B (radii of same circle)
O'O = OO' (common side)
ΔOAO' ΔOBO' (SSS congruency)
AOO' = BOO'
AOM = BOM ......(i)
Now in ΔAOM and ΔBOM we have
OA = OB (radii of same circle)
AOM = BOM (from (i))
OM = OM (common side)
ΔAOM ΔBOM (SAS congruency)
AM = BM and AMO = BMO
But
AMO + BMO = 180°
2AMO = 180°
AMO = 90°
Thus, AM = BM and
AMO = BMO = 90°
Hence OO' is the perpendicular bisector of AB.

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