M and N are points on opposite sides AD and BC of a parallelogram ABCD such that MN passes through the point of intersection O of its diagonals AC and BD. Show that MN is bisected at O.
Given: A parallelogram ABCD
To prove: MN is bisected at O
Proof:
In ∆OAM and ∆OCN,
OA = OC
(D iagonals of parallelogram bisect each other)
∠AOM = ∠CON (Vertically opposite angles)
∠MAO = ∠OCN (Alternate interior angles)
∴ By ASA congruence criteria,
∆OAM ≅ ∆OCN
⇒ OM = ON (CPCT)
Hence, MN is bisected at O.