E đối xứng D qua C \(\Rightarrow C\) là trung điểm DE
\(\Rightarrow\overrightarrow{DC}=\overrightarrow{CE}\Rightarrow\overrightarrow{CE}=\overrightarrow{AB}\)
Ta có:
\(\overrightarrow{AB}+\overrightarrow{AD}=\overrightarrow{CE}+\overrightarrow{BC}=\overrightarrow{BE}\) (đpcm)
b.
Đặt \(\overrightarrow{u}=\overrightarrow{DB}+\overrightarrow{DE}=\overrightarrow{DA}+\overrightarrow{AB}+2\overrightarrow{DC}=\overrightarrow{DA}+3\overrightarrow{DC}\)
\(\Rightarrow\left|\overrightarrow{u}\right|^2=\left(\overrightarrow{DA}+3\overrightarrow{DC}\right)^2=DA^2+9DC^2+6\overrightarrow{DA}.\overrightarrow{DC}=a^2+9a^2=10a^2\)
\(\Rightarrow\left|\overrightarrow{u}\right|=a\sqrt{10}\)
