Question

In the following figure, two triangles ABC and DBC are formed on same base BC. If AD, BC intersect at point O then show that (ar.(ABC))/(ar.(DBC)) = AO/DO

Answer 1

Given : ∆ABC and ∆DBC are two triangles on same base BC. AD, BC intersects each other at point O.

To prove : (ar.(ABC))/(ar.(DBC)) = AO/DO

Construction : From vertices A and D draw AE ⊥ BC and DE ⊥ BC respectively.

Proof : From vertices A and D, AE ⊥ BC and DF ⊥ BC.

∴ ∠AEO = ∠DFO = 90°

In right angled ∆AEO and ∆DFO

∠AEO = ∠DFO (each 90°)

∠AOE = ∠DOF (vertically opposite angles)

By A-A Similarity criterion

∆AEO – ∆DFO

⇒ AE/DF = AO/DO

Now area of ∆ABC = 1/2 BC × AE

and area of ∆DBC = 1/2 BC × DF

from equation (i) and (ii)

(ar.(ABC))/(ar.(DBC)) = AO/DO