Question

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

Answer 1

To prove: `(ar(ΔABC))/(ar(ΔDBC)) = (AO)/(DO)`

Construction: Draw AE ⊥ BC and DF ⊥ BC.

Proof: In ΔAOE and ΔDOF,

∠AOE = ∠DOF   ...(Vertically opposite angles)

∠AEO ∼ ∠DFO   ...(Each 90°)

ΔAOE ∼ ΔDOF   ...(By AA similarity)

`(AO)/(DO) = (AE)/(DF)`   ...(i)

Now, `(ar(ΔABC))/(ar(ΔDBC))`

= `(1/2 xx BC xx AE)/(1/2 xx BC xx DF)`

= `(AE)/(DF)`

`(ar(ΔABC))/(ar(ΔDBC)) = (AO)/(DO)`   ...[From equation (i)]

Hence Proved.