ABCD is a parallelogram with diagonals AC and BC intersecting O Prove that triangle AOB is congruent to triangle DOC

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.

maybe showing the graph would

step-by-step explanation:

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