Firstly draw two circles with center O and O’ such that they intersect at A and B.
Draw a line PQ parallel to OO’.
In the circle with center O, we have:
OP and OB are the radii of the circle. PB is the chord with OM as its perpendicular bisector.
i.e. BM = MP ………………………(1)
In the circle with center O’, we have:
O’B and O’Q are the radii of the circle. BQ is the chord with O’N as its perpendicular bisector.
i.e. BN = NQ ………………………(2)
From (1) and (2), we have:
BM + BN = MP + NQ
⇒ (BM + BN) + (BM + BN) = (BM + BN) + (MP + NQ)
⇒ 2(BM + BN) = (BM + BN) + (MP + NQ)
⇒ 2(OO’) = (BM + MP) + (BN + NQ)
⇒ 2(OO’) = BP + BQ
⇒ 2(OO’) = PQ
Hence, proved.