Question

If two circles intersect, prove that the line joining the centres is the perpendicular bisector of the common chord.

Answer 1

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 $OA,OB,O^{\prime }A$ and $O^{\prime }B$

In $\Delta OA{O}^{\prime }and\Delta OB{O}^{\prime }$ , we have

$OA=OB$ (radii of same circle)

$O^{\prime }A=O^{\prime }B$ (radii of same circle)

$O^{\prime }O=O{O}^{\prime }$ (common side)

Or, $\Delta OA{O}^{\prime }\cong \Delta OB{O}^{\prime }$ (SSS congruency)

$\angle AO{O}^{\prime }=\angle BO{O}^{\prime }$

$\angle AOM=\angle BOM$......(i)

Now in $\Delta AOMand\Delta BOM$ we have

$OA=OB$ (radii of same circle)

$\angle AOM=\angle BOM$ (from (i))

$OM=OM$ (common side)

Or, $\Delta AOM\cong \Delta BOM$ (SAS congruncy)

Or, $AM=BMand\angle AMO=\angle BMO$

But

$\angle AMO+\angle BMO={180}^{\circ }$

Or, $2\angle AMO={180}^{\circ }$

Or, $\angle AMO={90}^{\circ }$

Thus, $AM=BM$ and $\angle AMO=\angle BMO={90}^{\circ }$

Hence, OO' is the perpendicular bisector of AB.