Given: ABCD is a parallelogram with diagonals AC and BD intersecting at O.
To prove: ∆AOB ≅ ∆DOC
Proof: Since ABCD is a parallelogram.
Therefore, AB || CD and AB = CD, AD || BC and AD = BC
Now, AB || CD and transversal AC intersects them a A and C respectively.
Therefore, Angle BAC = Angle DCA [Since alternate interior angles are equal]
or, Angle BAO = Angle DCO
Now, in ∆AOB and ∆DOC,
Angle BAO = Angle DCO [proved above]
AB = CD [opposite sides of a parallelogram are equal]
Angle AOB = Angle COD [vertically opposite angles are equal]
So, by ASA congruency,
∆AOB ≅ ∆DOC.