Two circles whose centres are O and O′ intersect at P. Through P, a line parallel to OO′, intersecting the circle at C and D is drawn as shown. Then CD=2OO′.
Draw perpendiculars OA and O′B to CD from O and O′, respectively.
Since OA⊥CD, OA bisects the chord CP. (∵ Perpendicular from the centre of the chord bisects the chord)
or CP=2 AP - - - - - (i)
Similarly since O′B⊥PD, we must have BP=12PD
or PD=2 BP - - - - - (ii)
⟹CD=2AP+ABP=2(AP+BP) [from (i) and (ii)]
⟹CD=2(AB) - - - - - (iii)
In quadrilateral ABO′O, OO′∥AB and ∠OAB=∠O′BA=90∘
Thus ABO′O is a rectangle.
Therefore the given statement is true.