Giải thích các bước giải:
Ta có:
$\begin{array}{l}
+ )\overrightarrow {AD} = \dfrac{1}{2}\left( {\overrightarrow {AI} + \overrightarrow {AC} } \right)\\
= \dfrac{1}{2}\overrightarrow {AI} + \dfrac{1}{2}\overrightarrow {AC} \\
= \dfrac{1}{2}.\dfrac{1}{2}\overrightarrow {AB} + \dfrac{1}{2}\overrightarrow {AC} \\
= \dfrac{1}{4}\overrightarrow {AB} + \dfrac{1}{2}\overrightarrow {AC} \\
+ )\overrightarrow {ND} = \dfrac{1}{2}\left( {\overrightarrow {NI} + \overrightarrow {NC} } \right)\\
= \dfrac{1}{2}\overrightarrow {NI} + \dfrac{1}{2}\overrightarrow {NC} \\
= \dfrac{1}{2}\left( {\overrightarrow {NB} + \overrightarrow {BI} } \right) + \dfrac{1}{2}.\dfrac{1}{3}\overrightarrow {BC} \\
= \dfrac{1}{2}\overrightarrow {NB} + \dfrac{1}{2}\overrightarrow {BI} + \dfrac{1}{6}\overrightarrow {BC} \\
= \dfrac{1}{2}.\left( {\dfrac{{ - 2}}{3}\overrightarrow {BC} } \right) + \dfrac{1}{2}.\left( {\dfrac{{ - 1}}{2}\overrightarrow {AB} } \right) + \dfrac{1}{6}\overrightarrow {BC} \\
= - \dfrac{1}{6}\overrightarrow {BC} - \dfrac{1}{4}\overrightarrow {AB} \\
= - \dfrac{1}{6}\left( {\overrightarrow {AC} - \overrightarrow {AB} } \right) - \dfrac{1}{4}\overrightarrow {AB} \\
= - \dfrac{1}{{12}}\overrightarrow {AB} - \dfrac{1}{6}\overrightarrow {AC}
\end{array}$
$\begin{array}{l}
\Rightarrow \overrightarrow {AD} = - 3\overrightarrow {ND} \\
\Rightarrow AD = 3ND\\
\Rightarrow AN = 4ND
\end{array}$
Ta có điều phải chứng minh
