Question

The base BC of triangle ABC is divided at D so that BD = `1/2`DC.
Prove that area of ΔABD = `1/3` of the area of ΔABC.

Answer 1

In ΔABC, ∵ BD = `1/2"DC" ⇒ "BD"/"DC" = 1/2`

∴ Ar.( ΔABD ) : Ar.( ΔADC ) = 1:2

But Ar.( ΔABD ) + Ar.( ΔADC ) = Ar.( ΔABC )

Ar.( ΔABD ) + 2Ar.( ΔABD ) = Ar.( ΔABC )

3 Ar.( ΔABD ) = Ar.( ΔABC )

Ar.( ΔABD ) = `1/3`Ar.( ΔABC )