the point D divides the side BC of triangle ABc in the ratio m:n prove that ar(triangle ABD):ar(triangle ADC)=m:n
given: ABC is a triagle and d divides the BC in ratio m;n
construction: draw AE perpendicular to BC
proof: let BD = mx and dc=nx.
ar[ABD]=1/2 bh= 1/2 x BD x AE = 1/2 x AE x [mx] ...................... 1
ar[ADC]=1/2bh = 1/2 x DC x AE= 1/2 x AE x [nx] ...................... 2
from 1 and 2 we get,
ratio of ar[ABD] : ar[ADC] = m:n
construction: draw AE perpendicular to BC
proof: let BD = mx and dc=nx.
ar[ABD]=1/2 bh= 1/2 x BD x AE = 1/2 x AE x [mx] ...................... 1
ar[ADC]=1/2bh = 1/2 x DC x AE= 1/2 x AE x [nx] ...................... 2
from 1 and 2 we get,
ratio of ar[ABD] : ar[ADC] = m:n
