the base BC of a triangle ABC is divided at D so that BD=1/3 BC .prove that ar(triangle ABD)=1/2 ar (triangle ADC)

Answer 1

isn't the height of the triangle the same, so the only thing different about the two new triangles is the base?

Area ABD=1/2 Base*height= 1/2 1/3BC*h
area ADC= 1/2 Base*height= 1/2 2/3BC*h

dividefirst area by second

ratio of areas= 1/2

Answer 2

Not informative concept vise.Worst answer everrrrr.Shame of the person who gave this solution.