Diagonals AC and BD of a trapezium ABCD with 
intersect each other at the point O. Using similarity criterion for two triangles, show that 
.
We have,
ABCD is a trapezium with AB || DC
In ΔAOB and ΔCOD <AOB = <COD (Vertically opposite angle)
<OAB = <OCD (Alternate interior angle)
Then, ΔAOB ~ΔCOD (By AA similarity)
So, OA/OC = OB/OD(Corresponding parts of similar triangle are proportional)
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