a.
\(2\overrightarrow{IB}+\overrightarrow{IA}=\overrightarrow{0}\Rightarrow\overrightarrow{IB}=\dfrac{1}{3}\overrightarrow{AB}\) , tương tự \(\overrightarrow{CK}=\dfrac{1}{3}\overrightarrow{CD}\);
\(\overrightarrow{EB}=\dfrac{3}{5}\overrightarrow{CB}\) ; \(\overrightarrow{AF}=\dfrac{3}{5}\overrightarrow{AD}\)
Ta có:
\(\overrightarrow{IK}=\overrightarrow{IB}+\overrightarrow{BC}+\overrightarrow{CK}=\dfrac{1}{3}\overrightarrow{AB}+\overrightarrow{BC}+\dfrac{1}{3}\overrightarrow{CD}\)
\(=\dfrac{1}{3}\overrightarrow{AD}+\dfrac{1}{3}\overrightarrow{DC}+\dfrac{1}{3}\overrightarrow{CB}+\overrightarrow{BC}+\dfrac{1}{3}\overrightarrow{CD}\)
\(=\dfrac{1}{3}\overrightarrow{AD}+\dfrac{2}{3}\overrightarrow{BC}\)
\(\Rightarrow\overrightarrow{IK};\overrightarrow{BC};\overrightarrow{AD}\) đồng phẳng
Tương tự:
\(\overrightarrow{EF}=\overrightarrow{EB}+\overrightarrow{BA}+\overrightarrow{AF}=\dfrac{3}{5}\overrightarrow{CB}+\overrightarrow{BA}+\dfrac{3}{5}\overrightarrow{AD}\)
\(=\dfrac{3}{5}\overrightarrow{CD}+\dfrac{3}{5}\overrightarrow{DB}+\overrightarrow{BA}+\dfrac{3}{5}\overrightarrow{AD}=\dfrac{3}{5}\overrightarrow{CD}+\overrightarrow{BA}+\dfrac{3}{5}\overrightarrow{AB}\)
\(=\dfrac{2}{5}\overrightarrow{BA}+\dfrac{3}{5}\overrightarrow{CD}\)
\(\Rightarrow\overrightarrow{BA};\overrightarrow{EF};\overrightarrow{CD}\) đồng phẳng
b.
Kéo dài IF cắt BD tại M, đặt \(\overrightarrow{MB}=x.\overrightarrow{BD}\)
\(\overrightarrow{MI}=\overrightarrow{MB}+\overrightarrow{BI}=x.\overrightarrow{BD}+\dfrac{1}{3}\overrightarrow{BA}\)
\(\overrightarrow{MF}=\overrightarrow{MD}+\overrightarrow{DF}=\left(x+1\right)\overrightarrow{BD}+\dfrac{2}{5}\overrightarrow{DA}=\left(x+1\right)\overrightarrow{BD}+\dfrac{2}{5}\left(\overrightarrow{DB}+\overrightarrow{BA}\right)\)
\(=\left(x+\dfrac{3}{5}\right)\overrightarrow{BD}+\dfrac{2}{5}\overrightarrow{BA}\)
Do M,I,F thẳng hàng \(\Rightarrow\dfrac{x+\dfrac{3}{5}}{x}=\dfrac{\dfrac{2}{5}}{\dfrac{1}{3}}\Rightarrow x=3\) \(\Rightarrow\overrightarrow{MB}=3\overrightarrow{BD}\)
\(\Rightarrow\overrightarrow{ME}=\overrightarrow{MB}+\overrightarrow{BE}=3\overrightarrow{BD}+\dfrac{3}{5}\overrightarrow{BC}=\dfrac{3}{5}\left(5\overrightarrow{BD}+\overrightarrow{BC}\right)\)
\(\overrightarrow{MK}=\overrightarrow{MB}+\overrightarrow{BC}+\overrightarrow{CK}=3\overrightarrow{BD}+\overrightarrow{BC}+\dfrac{1}{3}\overrightarrow{CD}=3\overrightarrow{BD}+\overrightarrow{BC}+\dfrac{1}{3}\left(\overrightarrow{CB}+\overrightarrow{BD}\right)\)
\(=\dfrac{10}{3}\overrightarrow{BD}+\dfrac{2}{3}\overrightarrow{BC}=\dfrac{2}{3}\left(5\overrightarrow{BD}+\overrightarrow{BC}\right)\)
\(\Rightarrow\overrightarrow{ME}=\dfrac{9}{10}\overrightarrow{MK}\Rightarrow\) M, E, K thẳng hàng
\(\Rightarrow\) 2 đường thẳng IF và EK cắt nhau tại M
\(\Rightarrow I;E;K;F\) đồng phẳng
