Xét tg ABD và tg BCD có đường cao từ D->AB = đường cao từ B->CD nên
\(\frac{S_{ABD}}{S_{BCD}}=\frac{AB}{CD}=\frac{4}{5}\)
\(S_{ABCD}=S_{ABD}+S_{BCD}\)
Chia \(S_{ABD}\) thành 4 phần bằng nhau thì \(S_{BCD}\) là 5 phần như thế
\(\Rightarrow\frac{S_{ABD}}{S_{ABCD}}=\frac{S_{ABD}}{S_{ABD}+S_{BCD}}=\frac{4}{4+5}=\frac{4}{9}\Rightarrow S_{ABD}=\frac{4xS_{ABCD}}{9}\)
\(\Rightarrow\frac{S_{BCD}}{S_{ABCD}}=\frac{5}{9}\Rightarrow S_{BCD}=\frac{5xS_{ABCD}}{9}\)
Ta có \(\frac{AM}{MD}=2\Rightarrow\frac{AM}{AD}=\frac{2}{3};\frac{NC}{BN}=\frac{3}{2}\Rightarrow\frac{NC}{BC}=\frac{3}{5}\)
Xét tg ABM và tg ABD có chung đường cao từ B->AD nên
\(\frac{S_{ABM}}{S_{ABD}}=\frac{AM}{AD}=\frac{2}{3}\Rightarrow S_{ABM}=\frac{2xS_{ABD}}{3}=\frac{2}{3}x\frac{4xS_{ABCD}}{9}=\frac{8xS_{ABCD}}{27}\)
Xét tg CDN và tg BCD có chung đường cao tư D->BC nên
\(\frac{S_{CDN}}{S_{BCD}}=\frac{CN}{BC}=\frac{3}{5}\Rightarrow S_{CDN}=\frac{3}{5}xS_{BCD}=\frac{3}{5}x\frac{5xS_{ABCD}}{9}=\frac{S_{ABCD}}{3}\)
Ta có
\(S_{BMDC}=S_{ABCD}-S_{ABM}=S_{ABCD}-\frac{8xS_{ABCD}}{27}=\frac{19xS_{ABCD}}{27}\)
\(S_{ABND}=S_{ABCD}-S_{CDN}=S_{ABCD}-\frac{S_{ABCD}}{3}=\frac{2xS_{ABCD}}{3}\)
\(\Rightarrow S_{BMDC}-S_{ABND}=\frac{19xS_{BCD}}{27}-\frac{2xS_{ABCD}}{3}=\frac{S_{ABCD}}{27}=72\Rightarrow S_{ABCD}=27x72=1944cm^2\)
