
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.