Two circles with centres O and O' intersect at two points A and B. A line PQ is drawn parallel to OO' through A (or B) intersecting the circles at P and Q. prove that PQ = 2 OO'.
Given, draw two circles having centres O and O' intersect at points A and B.
Also, draw line PQ parallel to OO'
Construction : Join OO', OP, O'Q, OM and O'N
To prove that PQ = 2 OO’
In ΔOPB, BM = MP [OM is the perpendicular bisector of PB]
And in ΔO′BQ, BN = NQ [O'N is the perpendicular bisector of BQ]
∴ BM + BN = PM + NQ
⇒ 2(BM + BN) = BM + BN + PM + NQ [adding both sides BM + BN]
⇒ 2 OO' = (BM + MP) + (BN + NQ)
⇒ 2 OO' = (PB + (BQ)
⇒ 2 OO' = PQ, hence proved.