The diagonals of a quadrilateral ABCD intersect each other at point O such that $AO over BO equals CO over DO.$ Show that ABCD is a trapezium.

Given a quadrilateral ABCD in which AC and BD are diagonals, which intersect each other at O.
To Prove : ABCD is a trapezium such that AB || DC.

Const : Draw a line OM || AB.
Proof: In ∆ADB, we have OM || AB.
Therefore, by using Basic proportionality theorem, we have

DM over MA equals DO over OB

rightwards double arrow space space space space space space space space space space space space AM over DM equals OB over OD space space space space space space space... left parenthesis straight i right parenthesis

[Taking reciprocals of both sides]
It is given that,

space space space AO over BO equals CO over DO

rightwards double arrow

AO over OC equals OB over OD

...(ii)
Comparing (i) and (ii), we get

AM over DM equals OA over OC

Therefore, by using converse of basic proportionality theorem, we have
OM || DC
But OM || AB (by construction)
⇒ AB || DC
Hence, ABCD is a trapezium.

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