Đặt AD=a =>DC=2a
Ta có : AD2+DE2=AE2(theo py ta go)
=> AE=\(\frac{\sqrt{10}a}{3}\) =>CosDAE=\(\frac{a}{\frac{\sqrt{10}a}{3}}=\frac{3}{\sqrt{10}}\)
Gọi \(\overrightarrow{n_{AD}}\left(m,n\right)\)
CosDAE=\(\frac{\left|\overrightarrow{n_{AD}}\cdot\overrightarrow{n_{AE}}\right|}{\left|\overrightarrow{n_{AD}}\right|\left|\overrightarrow{n_{AE}}\right|}\)
\(\Rightarrow\frac{3}{\sqrt{10}}=\frac{\left|2m+\frac{2n}{3}\right|}{\sqrt{m^2+n^2}\sqrt{4+\frac{4}{9}}}\)
\(\Rightarrow4n^2-3mn=0\Rightarrow n\left(4n-3m\right)=0\)
TH1:chọn n=0=>m=1
=>D(1,0)
Ta có :\(\overrightarrow{EC}=-5\overrightarrow{ED}\)
\(\Rightarrow C\left(5,-6\right)\)=>I(3,-5/2)=>B(...)
TH2: chọn n=3=>m=4
=>AD:4x+3y-7=0
Do AD vuông góc vs DC=>DC:3x-4y-9=0
=>D(11/5,-3/5)
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