PUMPA - SMART LEARNING
எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்
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\(\frac{AO}{BO} = \frac{CO}{DO}\)
\(\angle AOB \sim \triangle COD\)
\(\frac{BO}{DO}\)
[Since corresponding sides of similar triangles]
[Since they are alternate angles]
\(AB\) is parallel to \(CD\)
\(\angle DCO\)
\(\frac{CO}{DO}\)
\(\angle OBA\)
\(ABCD\) 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, \(\angle OAB\) \(=\) 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 \(\frac{AO}{BO}\) \(=\) .
Hence, the required condition is proved.
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