Question

Two circles with centres O and O’ intersect each other at A and B. Prove that the centre line OO’ is the perpendicular bisector of the common chord AB

Answer 1

Answer:

Let two circles with centres O and O' intersect at two points A and B so that AB is the common chord of two circles and 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

′

andΔOBO

′

, we have

OA=OB (radii of same circle)

O

′

A=O

′

B (radii of same circle)

O

′

O=OO

′

(common side)

Or, ΔOAO

′

≅ΔOBO

′

(SSS congruency)

∠AOO

′

=∠BOO

′

∠AOM=∠BOM......(i)

Now in ΔAOMandΔBOM we have

OA=OB (radii of same circle)

∠AOM=∠BOM (from (i))

OM=OM (common side)

Or, ΔAOM≅ΔBOM (SAS congruncy)

Or, AM=BMand∠AMO=∠BMO

But

∠AMO+∠BMO=180

∘

Or, 2∠AMO=180

∘

Or, ∠AMO=90

∘

Thus, AM=BM and ∠AMO=∠BMO=90

∘

Hence, OO' is the perpendicular bisector of AB.