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# 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)/area ($$\triangle$$ DBC) = AO/DO.

Given, ABC and DBC are two triangles on the same base BC. ADintersects BC at O.

We have to prove: Area ($$\triangle$$ ABC)/Area ($$\triangle$$ DBC) = AO/DO

Let us draw two perpendiculars AP and DM on line BC.

We know that area of a triangle = $$\frac{1}{2}$$ × Base × Height
$${{ar(\triangle ABC) } \over {ar(\triangle DEF}} = {{{{1} \over {2}}BC(AP)}\over {{{1} \over {2}}BC(DM)}} ={{AP} \over {DM}}$$

In $$\triangle$$ APO and $$\triangle$$ DMO,
$$\angle$$ APO = $$\angle$$ DMO (Each 90°)
$$\angle$$ AOP = $$\angle$$ DOM
(Vertically opposite angles)
$$\triangle$$ APO ~ $$\triangle$$ DMO (AA similarity criterion)
$$\frac{AP}{DM} = \frac{AO}{DO}$$
$$\frac{Area (\triangle ABC)}{Area ( \triangle DBC)} = \frac{AO}{DO}$$ .