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Answer variants:
\(\frac{BO}{DO}\)
\(\frac{AO}{BO}\)
\(\angle OBA\)
\(\frac{AO}{BO} = \frac{CO}{DO}\)
\(\angle AOB \sim \triangle COD\)
\(AB\) is parallel to \(CD\)
[Since corresponding sides of similar triangles]
\(\angle OAB\)
[Since they are alternate angles]
\(ABCD\) is a trapezium. Here, the sides \(AB\) and \(CD\) are parallel to each other. Also, the diagonals intersect at \(O\). Prove that \(\frac{AO}{BO} = \frac{CO}{DO}\).
Let us look at the figure given below for a better understanding.
We already know that in \(\triangle ABC\) and \(\triangle COD\), . [Given]
This makes, \(=\) \(\angle DCO\), and \(=\) \(\angle ODC\).
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, .
Also, \(\frac{AO}{CO}\) \(=\) which can also be written as \(=\) \(\frac{CO}{DO}\).
Hence, the required condition is proved.
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