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Answer :

Given:

Two circles with centres O and O' intersect at two points M and N so that MN is the common chord of the two circles and OO' is the line segment joining the centres of the two circles. Let OO’ intersect MN at P.

To prove: OO' is the perpendicular bisector of MN.

Construction:

Draw line segments OM, ON, O'M and O'N.

Proof:

In \(\triangle{OMO'}\) and \(\triangle{ONO'}\), we have,

OM = ON ...(Radii of congruent circles)

O'M = O'N ...(Radii of congruent circles)

and OO' = O'O ...(Common)

By SSS criterion, we get,

\(\triangle{OMO'}\) \(\displaystyle \cong\) \(\triangle{ONO'}\)

Hence, \(\angle{MOO'}\) = \(\angle{NOO'}\) ...(By CPCT)

\(\therefore \) \(\angle{MOP}\) = \(\angle{NOP}\) ...(i)(\(\because \) \(\angle{MOO'}\) = \(\angle{MOP}\), \(\angle{NOO'}\) = \(\angle{NOP}\))

In \(\triangle{MOP}\) and \(\triangle{NOP}\), we have,

OM = ON ...(Radii of congruent circles)

OM = OM ...(Common)

and \(\angle{MOP}\) = \(\angle{NOP}\) ...(from eq. (i))

By SAS criterion, we get,

\(\triangle{MOP}\) \(\displaystyle \cong\) \(\triangle{NOP}\)

Hence, MP = NP ...(By CPCT)

And \(\angle{MPO}\) = \(\angle{NPO}\) ...(ii)

But \(\angle{MPO}\) + \(\angle{NPO}\) = \(180^\circ\)

(Since, MPN is a straight line)

\(\therefore \) 2 \(\angle{MPO}\) = \(90^\circ\) ...(from (ii))

\(\therefore \) \(\angle{MPO}\) = \(90^\circ\)

We have,

MP = NP and \(\angle{MPO}\) = \(\angle{NPO}\) = \(90^\circ\)

Hence, it is proved that OO' is the perpendicular bisector of MN.

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