Given:The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO i.e., AO/CO = BO/DO
To Prove: ABCD is a trapezium
Construction: Draw OE | DC such that E lies on BC.
Proof: In △BDC,
By Basic Proportionality Theorem, BO/OD = BE/EC (1)
But, AO/CO = BO/DO (Given) (2)
∴From (1) and (2) AO/CO = BE/EC
Hence, By Converse of Basic Proportionality Theorem, OE | AB
Now Since, AB | OE | DC
∴AB | DC
Hence, ABCD is a trapezium.