Question

In the given figure, ΔABC and ΔDBC have the same base BC. If AD and BC intersect at O, prove that `(ar(ΔABC))/(ar(ΔDBC)) = (AO)/(DO)`.

Answer 1

Given: Triangles ΔABC and ΔDBC have the same base BC. AD and BC meet at O.

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

Proof [Step-wise]:

1. Recall: Area of a triangle = `1/2` × base × height. 

Hence, if two triangles have the same altitude (height) then their areas are proportional to their bases.

2. Consider triangles ABO and DBO. Their bases AO and DO lie on the same line AD, and both triangles have the same altitude from B to line AD.

Therefore `(ar(ABO))/(ar(DBO)) = (AO)/(DO)`.

3. Similarly, for triangles ACO and DCO the bases are AO and DO on AD and both have the same altitude from C to AD. 

Hence `(ar(ACO))/(ar(DCO)) = (AO)/(DO)`.

4. Add the two equalities from steps 2 and 3:

`(ar(ABO) + ar(ACO))/(ar(DBO) + ar(DCO)) = (AO)/(DO)`

5. Note that ar(ABO) + ar(ACO) = ar(ABC) and ar(DBO) + ar(DCO) = ar(DBC).

Substituting gives `(ar(ABC))/(ar(DBC)) = (AO)/(DO)`.

Hence ar(ΔABC) : ar(ΔDBC) = AO : DO, as required.