Giải thích các bước giải:
Ta có : $A, I, B'$ thẳng hàng
$\to $Áp dụng định lý Menelauyt ta được :
$\dfrac{AC}{AA'}.\dfrac{IA'}{IB}.\dfrac{B'B}{B'C}=1$
$\to\dfrac{IA'}{IB}=\dfrac{AA'}{AC}.\dfrac{B'C}{B'B}$
$\to\dfrac{IA'}{IB}=(1-m).\dfrac{1}{1-n}$
$\to IA'=(1-m).\dfrac{1}{1-n}.IB$
$\to IA'+IB=(1+(1-m).\dfrac{1}{1-n}).IB$
$\to A'B=\dfrac{2-m-n}{1-n}.IB$
$\to \dfrac{BI}{BA'}=\dfrac{1-n}{2-m-n}$
$\to \vec{CI}=\vec{CB}+\vec{BI}=\vec{b}+\dfrac{1-n}{2-m-n}.\vec{BA'}$
$\to \vec{CI}=\vec{b}+\dfrac{1-n}{2-m-n}.(\vec{BC}+\vec{CA'})$
$\to \vec{CI}=\vec{b}+\dfrac{1-n}{2-m-n}.(-\vec{b}+m\vec{a})$
$\to \vec{CI}=\vec{b}.\dfrac{1-m}{2-m-n}+\dfrac{m(1-n)}{2-m-n}\vec{a}$
Mà $\vec{CI}=m\vec a+n\vec b$
$\to \begin{cases}\dfrac{m(1-n)}{2-m-n}=m\\\dfrac{1-m}{2-m-n}=n \end{cases}$
$\to \begin{cases}m(1-n)=m(2-m-n)\\1-m=(2-m-n)n \end{cases}$
$\to \begin{cases}m^2-m=0\\1-m-2n+mn+n^2=0\end{cases}$
$+) m=0\to n^2-2n+1=0\to n=1$
$+) m=1\to n^2-n=0\to n\in\{0,1\}$
