#### 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 for 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) concept: Areas of Similar Triangles. For the course CBSE