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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 xx "Base" xx "Height"`
`:.(ar(triangleABC))/(ar(triangleDBC)) = (1/2 BC xx AP)/(1/2BC xx DM) = (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)`
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