Question

In the figure the point d divides the side BC of ∆ABC in the ratio m:n, prove that area of ∆ABD:area of ∆ADC=m:n

Answer 1

Your question figure is not visible so plzzzz do link if it's the same

Answer 2

Answer:

Area ABD / area ADC = [(½) m AD sin BDA] / [(½) n AD sin BDA] = m/n

Step-by-step explanation:

Area ABD = (½) x base (m) x height of BC from vertex A

Area ACD = (½) x base (n) x height of BC from vertex A