Solution :
Given : Two triangles ABC and DBC which stand on the same base BC but on the opposite sides of BC.
To Prove : \(area(\triangle ABC)\over area(\triangle DBC)\) = \(AO\over DO\)
Construction : Draw AE \(\perp\) BC and DF \(\perp\) BC.
Proof : In triangles AOE and DOF, we have :
\(\angle\) AEO = \(\angle\) DFO = 90
\(\angle\) AOE = \(\angle\) DOF (vertically opposite angles)
By AA similarity, we have :
\(\triangle\) AOE ~ \(\triangle\) DOF
So, \(AE\over DF\) = \(AO\over OD\) ……..(1)
Now, \(area(\triangle ABC)\over area(\triangle DBC)\) = \({1\over 2} \times BC \times AE\over {1\over 2}\times BC \times DF\) = \(AE\over DF\) = \(AO\over OD\)
\(\therefore\) \(area(\triangle ABC)\over area(\triangle DBC)\) = \(AO\over OD\).