\(ABCD\) is a trapezium. Here, the sides \(AB\) and \(CD\) are parallel to each other. Also, the diagonals intersect at \(O\). Verify that \(\frac{AO}{BO} = \frac{CO}{DO}\).
Let us look at the below figure for a better understanding.
In \(\triangle AOB\) and \(\triangle COD\), \(AB\) is parallel to \(CD\). [Given]
Thus, \(\angle OAB\ =\) (Alternate angles)
\(\angle OBA\ =\) (Alternate angles)
By similarity critertion, we have, "If two of one triangle are respectively equal to two of another triangle, then the two triangles are similar."
Thus, \(\triangle AOB \sim\) .
Also, \(\frac{AO}{CO}\ =\), which can also be written as
[Since corresponding sides of similar triangles]
Hence, the required condition \(\frac{AO}{BO} = \frac{CO}{DO}\) is proved.