Suppose line segments AB and CD intersect at O in such a way that AO = OD and OB = OC. Prove that AC = BD but AC may not be parallel to BD.
In ∆OAC and ∆ODB,
OA = OD | Given
OB = OC | Given
∠AOC = ∠DOB
| Vertically Opposite Angles
∴ ∆OAC ≅ ∆ODB | SAS Axiom
∴ AC = BD | C.P.C.T.
Also, ∠OAC = ∠ODB | C.P.C.T.
and ∠OCA = ∠OBD | C.P.C.T.
Thus ∠OAC may not be equal to ∠OBD and therefore, AC may not be parallel to BD
However, if OA = OC, then ∠OAC = ∠OCA
| Angles opposite to equal sides of ∆OAC But ∠OAC = ∠ODB
∴ ∠OCA = ∠ODB
But these angles form a pair of equal alternate angles
∴ AC || BD.