Cho hình vuông $ABCD$  $;E,F$ là các điểm xác định bởi $\overrightarrow {BE}  = \frac{1}{3}\overrightarrow {BC} ;\overrightarrow {CF}  =  - \frac{1}{2}\overrightarrow {CD} ,AE$ cắt $BF$ tại $I$. Chứng minh $\widehat {AIC} = 1v$.

Giả sử $\overrightarrow {BI}  = k\overrightarrow {BF}$.
Ta có :
$\begin{array}{l}
\overrightarrow {AE}  = \frac{2}{3}\overrightarrow {AB}  + \frac{1}{3}\overrightarrow {AC}  = \frac{2}{3}\overrightarrow {AB}  + \frac{1}{3}\left( {\overrightarrow {AB}  + \overrightarrow {AD} } \right)\\
 = \overrightarrow {AB}  + \frac{1}{3}\overrightarrow {AD}
\end{array}$
Lại có $\overrightarrow {AI}  = \overrightarrow {AB}  + \overrightarrow {BI}  = \overrightarrow {AB}  + k\overrightarrow {BF} $
$\begin{array}{l}
 = \overrightarrow {AB}  + k\left( {\overrightarrow {BC}  + \overrightarrow {CF} } \right)\\
 = \overrightarrow {AB}  + k\left( {\overrightarrow {AD}  + \frac{1}{2}\overrightarrow {AB} } \right)\\
 = \left( {1 + \frac{k}{2}} \right)\overrightarrow {AB}  + k\overrightarrow {AD}
\end{array}$         
Vì $\overrightarrow {AI}  \text {cùng hướng với}  \overrightarrow {AE} $ nên $\left( {1 + \frac{k}{2}} \right):1 = k:\frac{1}{3}$
$\begin{array}{l}
 \Rightarrow 1 + \frac{k}{2} = 3k\\
 \Rightarrow k = \frac{2}{5}
\end{array}$
Vậy $\overrightarrow {AI}  = \frac{6}{5}\overrightarrow {AB}  + \frac{2}{5}\overrightarrow {AD} $.
Đồng thời cũng có
$\begin{array}{l}
\overrightarrow {CI}  = \overrightarrow {AI}  - \overrightarrow {AC}  = \frac{6}{5}\overrightarrow {AB}  + \frac{2}{5}\overrightarrow {AD}  - \left( {\overrightarrow {AB}  + \overrightarrow {AD} } \right)\\
\,\,\,\,\, = \frac{1}{5}\overrightarrow {AB}  - \frac{3}{5}\overrightarrow {AD} \\
 \Rightarrow \overrightarrow {AI} .\overrightarrow {CI}  = \left( {\frac{6}{5}\overrightarrow {AB}  + \frac{2}{5}\overrightarrow {AD} } \right)\left( {\frac{1}{5}\overrightarrow {AB}  - \frac{3}{5}\overrightarrow {AD} } \right)
\end{array}$
   $ = \frac{6}{{25}}{a^2} - \frac{6}{{25}}{a^2} = 0$ ( $a$ là độ dài cạnh hình vuông)
   $ \Rightarrow AI \bot CI$ ( đpcm)

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