The daigonals of a quadrilateral abcd …

Given: The diagonals of a quadrilateral ABCD intersect each other at the point O such that {tex}\frac{{AO}}{{BO}} = \frac{{CO}}{{DO}}{/tex}
To prove: ABCD is trapezium.
Construction: Through O draw a line OE||BA intersecting AD at E.
Proof: In {tex}\triangle DBA{/tex}{tex}\because OE||BA{/tex}
 
{tex}\therefore \frac{{DO}}{{BO}} = \frac{{DE}}{{AE}} \Rightarrow \frac{{CO}}{{AO}} = \frac{{DE}}{{AE}}{/tex}
{tex}\because \frac{{AO}}{{BO}} = \frac{{CO}}{{DO}}\,[Given]{/tex}
{tex}\Rightarrow \frac{{DO}}{{BO}} = \frac{{CO}}{{AO}} \Rightarrow \frac{{AO}}{{CO}} = \frac{{AE}}{{DE}}{/tex}.........[Taking reciprocals]
{tex}\therefore {/tex}In {tex}\triangle ADC{/tex}
OE {tex}\parallel{/tex} CD ...........[By converse basic proportionality theorem]
But OE {tex}\parallel{/tex} BA
BA {tex}\parallel{/tex} CD........[By construction]
The quadrilateral ABCD is a Trapezium.