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Lời giải:
$\overrightarrow{AB}\parallel \overrightarrow{C'D'}$ và $|\overrightarrow{AB}|=|\overrightarrow{C'D'}|=a$ nên:
$\overrightarrow{AB}.\overrightarrow{C'D'}=a^2$
\(\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\Rightarrow\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a}\)
Theo Talet: \(\dfrac{A'K}{IK}=\dfrac{B'I}{A'D'}=\dfrac{1}{2}\Rightarrow A'K=\dfrac{2}{3}A'I\)
\(\Rightarrow\overrightarrow{A'K}=\dfrac{2}{3}\overrightarrow{A'I}=\dfrac{2}{3}\left(\overrightarrow{A'B'}+\overrightarrow{B'I}\right)=\dfrac{2}{3}\left(\overrightarrow{A'B'}+\dfrac{1}{2}\overrightarrow{B'C'}\right)\)
\(=\dfrac{2}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{BC}=\dfrac{2}{3}\overrightarrow{a}+\dfrac{1}{3}\left(\overrightarrow{b}-\overrightarrow{a}\right)=\dfrac{1}{3}\overrightarrow{a}+\dfrac{1}{3}\overrightarrow{b}\)
\(\Rightarrow\overrightarrow{DK}=\overrightarrow{DD'}+\overrightarrow{D'A'}+\overrightarrow{A'K}=\overrightarrow{AA'}-\overrightarrow{BC}+\overrightarrow{A'K}\)
\(=\overrightarrow{c}-\left(\overrightarrow{b}-\overrightarrow{a}\right)+\dfrac{1}{3}\overrightarrow{a}+\dfrac{1}{3}\overrightarrow{b}\)
\(=\dfrac{4}{3}\overrightarrow{a}-\dfrac{2}{3}\overrightarrow{b}+\overrightarrow{c}\)
\(\overrightarrow{BD'}=\overrightarrow{BA}+\overrightarrow{AD}+\overrightarrow{DD'}=-\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{AA'}\)
\(=-\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\)
\(\overrightarrow{A'C}=\overrightarrow{A'A}+\overrightarrow{AC}=-\overrightarrow{AA'}+\overrightarrow{AB}+\overrightarrow{AD}=\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c}\)
\(\overrightarrow{BD}.\overrightarrow{D'C}=\overrightarrow{BD}\left(\overrightarrow{D'D}+\overrightarrow{DC}\right)=\overrightarrow{BD}.\overrightarrow{D'D}+\overrightarrow{BD}.\overrightarrow{DC}\)
\(=\overrightarrow{BD}.\overrightarrow{DC}=-\overrightarrow{DB}.\overrightarrow{DC}=-a\sqrt{2}.a.cos45^0=-a^2\)