Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that AO/OC = OB/OD

In \(\triangle\)DOC and \(\triangle\)BOA,

AB || CD, thus alternate interior angles will be equal,
\(\angle\)CDO = \(\angle\)ABO

Similarly,

\(\angle\)DCO = \(\angle\)BAO

Also, for the two triangles \(\triangle\)DOC and \(\triangle\)BOA, vertically opposite angles will be equal;
\(\angle\)DOC = \(\angle\)BOA
Hence, by AAA similarity criterion,
\(\triangle\)DOC ~ \(\triangle\)BOA

Thus, the corresponding sides are proportional.
DO/BO = OC/OA
OA/OC = OB/OD

Hence, proved.