# In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that $$area(\triangle ABC)\over area(\triangle DBC)$$ = $$AO\over DO$$.

## 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, $$area(\triangle ABC)\over area(\triangle DBC)$$ = $${1\over 2} \times BC \times AE\over {1\over 2}\times BC \times DF$$ = $$AE\over DF$$ = $$AO\over OD$$

$$\therefore$$     $$area(\triangle ABC)\over area(\triangle DBC)$$ = $$AO\over OD$$.