Rysunek pomocniczy w załączniku.
[tex]|AD| : |DC| = 1:2\quad\implies\quad \overrightarrow{AD}=\frac13\overrightarrow{AC}\ \ \wedge\ \ \overrightarrow{DC} =\frac23\overrightarrow{AC}[/tex]
[tex]\overrightarrow{AD}=\overrightarrow{BD}-\overrightarrow{BA}\qquad\qquad \wedge \qquad\qquad\overrightarrow{DC}=\overrightarrow{BC}-\overrightarrow{BD}\\\\\frac13\overrightarrow{AC}= \overrightarrow{BD}-\overrightarrow{BA}\qquad\qquad \wedge \qquad\quad \frac23\overrightarrow{AC} =\overrightarrow{BC}-\overrightarrow{BD}\\\\\overrightarrow{AC}=3(\overrightarrow{BD}-\overrightarrow{BA})\qquad\quad \wedge \qquad\qquad\overrightarrow{AC}=\frac32(\overrightarrow{BC}-\overrightarrow{BD})[/tex]
[tex]3(\overrightarrow{BD}-\overrightarrow{BA})=\frac32(\overrightarrow{BC}-\overrightarrow{BD})\qquad/\cdot\frac23\\\\2(\overrightarrow{BD}-\overrightarrow{BA})=\overrightarrow{BC}-\overrightarrow{BD} \\\\ 2\overrightarrow{BD}+\overrightarrow{BD} = 2\overrightarrow{BA}+\overrightarrow{BC} \\\\ 3\overrightarrow{BD} = 2\overrightarrow{BA}+\overrightarrow{BC}\qquad/:3\\\\\overrightarrow{BD} = \frac23\overrightarrow{BA}+\frac13\overrightarrow{BC}[/tex]
Co należało wykazać.
