Given trapezium \(ABCD\) with \(AB\) parallel to \(CD\) and diagonals intersecting at \(O\). Show that \(\frac{AO}{BO}=\frac{CO}{DO}\).
Let us look at the figure given below for a better understanding.
In \(\triangle ABC\) and \(\triangle COD\), .
This makes, \(\angle OAB\) \(=\) and \(\angle ODC\)
Since
By similarity criterion, we have, "If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar."
Thus, \(\triangle AOB \sim\) .
Also, \(\frac{AO}{CO} =\) .
\(\frac{AO}{BO} =\)
Hence proved.