a: \(\left|\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{AC}\right|=2\cdot AC=2\cdot5=10\)

b: \(\left|\overrightarrow{AM}+\overrightarrow{AN}\right|=\left|\dfrac{\overrightarrow{AB}+\overrightarrow{AC}}{2}+\dfrac{\overrightarrow{AD}+\overrightarrow{AC}}{2}\right|\)

\(=\left|\dfrac{3\cdot\overrightarrow{AC}}{2}\right|=\dfrac{3}{2}AC=\dfrac{3}{2}\cdot5=\dfrac{15}{2}=7.5\)

\(\overrightarrow{AM}\cdot\overrightarrow{BC}=\overrightarrow{BC}\left(\overrightarrow{BM}-\overrightarrow{BA}\right)=\overrightarrow{BM}\cdot\overrightarrow{BC}-\overrightarrow{BC}\cdot\overrightarrow{BA}\)

\(=BM\cdot BC\cdot cos0^0=\dfrac{1}{2}\cdot a^2\cdot1=\dfrac{1}{2}a^2\)

\(\left|\overrightarrow{AM}+\overrightarrow{BC}\right|=\sqrt{AM^2+BC^2+2\cdot\dfrac{1}{2}a^2}\)

\(=\sqrt{\dfrac{1}{4}a^2+a^2+a^2+a^2}=\dfrac{\sqrt{13}}{2}\cdot a\)

\(\left\{{}\begin{matrix}AM=\sqrt{AB^2+BM^2}=3\sqrt{5}\\DM=\sqrt{CD^2+CM^2}=3\sqrt{5}\end{matrix}\right.\) \(\Rightarrow\) tam giác ADM cân tại M

Gọi F là trung điểm AD \(\Rightarrow ABMF\) là hình chữ nhật \(\Rightarrow MF=AB=6\)

Theo tính chất trọng tâm: \(GF=\dfrac{1}{3}MF=2\)

\(DF=\dfrac{1}{2}AD=3\)

Đặt \(T=\left|\overrightarrow{GD}\right|=\left|\overrightarrow{GF}+\overrightarrow{FD}\right|\)

\(\Rightarrow T^2=GF^2+FD^2+2\overrightarrow{GF}.\overrightarrow{DF}=GF^2+DF^2=2^2+3^2=13\) 

\(\Rightarrow\left|\overrightarrow{GD}\right|=\sqrt{13}\)

M đối xứng với gì qua D nhỉ?