# If two circles intersect at two points, prove that their center lie on the perpendicular bisector of the common chord.

Let two circles with centres O and O' respectively intersect at two points A and B
such that AB is the common chord of two circles and
OO' is the line segment joining the centres, as shown in the figure: Also, AB is the chord of the circle centered at O. Therefore, the perpendicular bisector of AB will pass through O.

Let OO' intersect AB at M. Also, draw line segments OA, OB, O'A and O'B. Now, In ΔOAO' and OBO',
OA = OB (radii of same circle)
O'A = O'B (radii of same circle)
O'O ( common side)
⇒ ΔOAO' ≅ ΔOBO' (Side-Side-Side 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 (Side-Angle-Side congruency)
⇒ AM = BM and ∠AMO = ∠BMO
But
∠AMO + ∠BMO = 180°
⇒ 2∠AMO = 180°
⇒ ∠AMO = 90°
Thus, AM = BM and ∠AMO = ∠BMO = 90°
Hence OO' is the perpendicular bisector of AB.

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