if the diagonals of a quadrilateral divide each other proportionally then it is a trapezium
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Let ABCD be a quadrilateral as shown in the above figure:
Diagonal AC and BD divide each other proportionately that is,
OC
AO
=
OD
BO
Therefore, the triangles containing these sides are equiangular that is △AOB and △COD and thus,
∠OAB=∠OCD
∠OBA=∠ODC (Alternate angles)
Therefore, DC∣∣AB
Hence, ABCD is a trapezium.
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