#### Question

In Figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that `(ar(ABC))/(ar(DBC)) = (AO)/(DO)`

#### Solution

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

We know that area of a triangle = 1/2 x Base x Height

`:.(ar(triangleABC))/(ar(triangleDBC)) = (1/2 BC XX AP)/(1/2BCxxDM)) = (AP)/(DM)`

In ΔAPO and ΔDMO,

∠APO = ∠DMO (Each = 90°)

∠AOP = ∠DOM (Vertically opposite angles)

∴ ΔAPO ∼ ΔDMO (By AA similarity criterion)

`:. (AP)/(DM) = (AO)/(DO)`

`=> (ar(triangleABC))/(ar(triangleDBC))=(AO)/(DO)`

Is there an error in this question or solution?

Solution In Figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that (ar(ABC))/(ar(DBC)) = (AO)/(DO) Concept: Areas of Similar Triangles.