\(\overrightarrow{BD'}=\overrightarrow{BA}+\overrightarrow{BC}+\overrightarrow{BB'}=\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\)
M thuộc AC nên \(\overrightarrow{AM}=x.\overrightarrow{AC}=x\left(\overrightarrow{AB}+\overrightarrow{BC}\right)=x\left(\overrightarrow{c}-\overrightarrow{a}\right)\)
N thuộc DC' nên \(\overrightarrow{DN}=y.\overrightarrow{DC'}=y.\left(\overrightarrow{DC}+\overrightarrow{CC'}\right)=y\left(\overrightarrow{b}-\overrightarrow{a}\right)\)
\(\overrightarrow{MN}=\overrightarrow{MA}+\overrightarrow{AD}+\overrightarrow{DN}=x\left(\overrightarrow{a}-\overrightarrow{c}\right)+\overrightarrow{c}+y\left(\overrightarrow{b}-\overrightarrow{a}\right)\)
\(=\left(x-y\right)\overrightarrow{a}+y.\overrightarrow{b}+\left(1-x\right)\overrightarrow{c}\)
MN song song BD' nên:
\(\dfrac{x-y}{1}=\dfrac{y}{1}=\dfrac{1-x}{1}\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\overrightarrow{MN}=\dfrac{1}{3}\left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)=\dfrac{1}{3}\overrightarrow{BD'}\)
\(\Rightarrow\dfrac{MN}{BD'}=\dfrac{1}{3}\)
