Question

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

Answer 1

consider height h of ABC
ar(ABC)=ar(ABD)+ar(ADC)
height is same area is based on base length
ar(ABD)/ar(ADC)=BD/DC= 1/2
HENCE PROVED!

Answer 2

If BD=1/3 BC then DC=2/3

BC/DC=1/3÷2/3=1/2.

hence D divides BC in the ratio of 1:2

Area(∆ABD)/area(∆ADc)=1/2

Therefore ar(∆ABD)=1/2ar(∆ADC)

.....hence proved