Answer :

Given: ABCD is a trapezium with AB || CD and the diagonals AC and BD intersect at 'O'.
To prove:


Proof:
In the figure consider the triangle OAB and OCD
∠DOC = ∠ΑOB (Vertically opposite angles are equal)
since AB || DC,
∠DCO = ∠OAB (Alternate angles are equal)
∴ By AA corollary of similar triangles.
∴ Δ OAB ∼ Δ OCB When the two triangle are similar, the side are proportionally.
⇒


Cross-multiplying, we get,

Hence proved.
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