Problem: Two circles \(\Gamma \) and \(\Sigma \), with centers O and O’, respectively, are such that O’ lies on \(\Gamma \). Let A be a point on \(\Sigma \), and let M be the midpoint of AO’. Let B be another point on \(\Sigma \), such that \(AB~||~OM \). Then prove that the midpoint of AB lies on \(\Gamma \).
Suppose AB intersects \(\Sigma \) at C. Join O’C. Suppose it intersects OM at D. Clearly in \(\Delta AO’C \) M is the midpoint of AO’ and DM is parallel to AC. Then D is the midpoint of O’C.
Now O’C is a chord of \(\Gamma \) and we have proved that D is the midpoint of it. Therefore we can say that OD is perpendicular to O’C.
Since AB is parallel to OD (OM), therefore as O’C is perpendicular to OD, therefore O’C is also perpendicular to AB. Since AB is a chord of circle \(\Sigma \) and O’C is a line from center perpendicular to the chord, hence it bisects are chord implying that C is the midpoint of AB.