1. Đặt [tex]AB'=y,B'C=b,CA'=x,A'B=a,BC'=z,C'A=c[/tex]
Ta có: [tex]\left\{\begin{matrix} \overrightarrow{AA'}=\frac{a}{a+x}\overrightarrow{AC}+\frac{x}{a+x}\overrightarrow{AB}\\ \overrightarrow{BB'}=\frac{y}{y+b}\overrightarrow{BC}+\frac{b}{y+b}\overrightarrow{BA}\\ \overrightarrow{CC'}=\frac{z}{z+c}\overrightarrow{CA}+\frac{c}{z+c}\overrightarrow{CB} \end{matrix}\right.\Rightarrow \overrightarrow{AA'}+\overrightarrow{BB'}+\overrightarrow{CC'}=(\frac{a}{a+x}-\frac{z}{z+c})\overrightarrow{AC}+(\frac{c}{z+c}-\frac{y}{y+b})\overrightarrow{CB}+(\frac{b}{y+b}-\frac{x}{x+a})\overrightarrow{BA}=(\frac{a}{a+x}-\frac{z}{z+c})(-\overrightarrow{BA}+\overrightarrow{CB})+(\frac{c}{z+c}-\frac{y}{y+b})\overrightarrow{CB}+(\frac{b}{y+b}-\frac{x}{x+a})\overrightarrow{BA}=(\frac{b}{y+b}-\frac{x}{x+a}+\frac{z}{z+c}-\frac{a}{a+x})\overrightarrow{BA}+(\frac{c}{z+c}-\frac{y}{y+b}+\frac{a}{a+x}-\frac{z}{z+c})\overrightarrow{CB}=\overrightarrow{0}\Rightarrow \left\{\begin{matrix} \frac{b}{y+b}-\frac{x}{x+a}+\frac{z}{z+c}-\frac{a}{a+x}=0\\ \frac{c}{z+c}-\frac{y}{y+b}+\frac{a}{a+x}-\frac{z}{z+c}=0 \end{matrix}\right. \Rightarrow \left\{\begin{matrix} \frac{1}{\frac{y}{b}+1}-\frac{1}{1+\frac{a}{x}}+\frac{1}{1+\frac{c}{z}}-\frac{1}{1+\frac{x}{a}}=0\\ \frac{1}{\frac{z}{c}+1}-\frac{1}{\frac{y}{b}+1}+\frac{1}{1+\frac{x}{a}}-\frac{1}{1+\frac{c}{z}}=0 \end{matrix}\right.[/tex]
Đặt [TEX]\frac{x}{a}=m,\frac{y}{b}=n,\frac{z}{c}=p[/TEX]
Ta có: [tex]\left\{\begin{matrix} \frac{1}{1+a}=1-\frac{1}{1+\frac{1}{a}}\\ \frac{1}{1+b}=1-\frac{1}{1+\frac{1}{b}}\\ \frac{1}{1+c}=1-\frac{1}{1+\frac{1}{c}} \end{matrix}\right.[/tex]
Từ trên ta có:[tex]\left\{\begin{matrix} \frac{1}{n+1}-(1-\frac{1}{m+1})+(1-\frac{1}{p+1})-\frac{1}{1+m}=0\\ \frac{1}{p+1}-\frac{1}{n+1}+\frac{1}{1+m}-(1-\frac{1}{p+1})=0 \end{matrix}\right.\Rightarrow \left\{\begin{matrix} \frac{1}{n+1}+\frac{1}{p+1}=1\\ \frac{1}{m+1}-\frac{1}{n+1}+\frac{2}{p+1}=1 \end{matrix}\right.\Rightarrow \left\{\begin{matrix} \frac{1}{n+1}+\frac{1}{p+1}=1\\ \frac{1}{m+1}+\frac{3}{p+1}=2 \end{matrix}\right.\Rightarrow n+p+2=np+n+p+1\Rightarrow np=1[/tex]
Theo định lí Ceva thì [TEX]mnp=1 \Rightarrow m=1[/TEX]
Thay lại ta có n = p = 1. Từ đó ta có đpcm.