giúp mk vssss

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
vì\ ABCD\ là\ hình\ vuông\ nên\ các\ góc=90\ độ,\ độ\ dài\ đường\ chéo\ a\sqrt{2}\\
\overrightarrow{AB} .\overrightarrow{AD} \ =|\overrightarrow{AB} |.|\overrightarrow{AD} |.cos( 90\ độ\ ) =a.a.0=0\\
sai\\
b,\ \overrightarrow{AC} .\overrightarrow{AB} \ =|\overrightarrow{AC} |.|\overrightarrow{AB} |.cos( 45\ độ) =\frac{\sqrt{2}}{2} a.a\sqrt{2\ }
\end{array}$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
=a^{2} \ mệnh\ đề\ sai\\
\end{array}$
$\displaystyle c,\ \overrightarrow{AB} +\overrightarrow{AD\ }$
$\displaystyle =\overrightarrow{DB}$
$\displaystyle \overrightarrow{BD} +\ \overrightarrow{BC} =\overrightarrow{DC}$ nhân vế trái với vế trái vế phải với vế phải ta có
$\displaystyle \overrightarrow{( AB} +\overrightarrow{AD}$).($\displaystyle \overrightarrow{BD} +\overrightarrow{BC}$)=($\displaystyle \overrightarrow{DB}$ .$\displaystyle \overrightarrow{DC}$)=|$\displaystyle \overrightarrow{DB|}$.$\displaystyle \overrightarrow{|DC\ }$|.$\displaystyle \cos( 90\ độ\ )$ =$\displaystyle a\sqrt{2}$ .$\displaystyle a\sqrt{2}$ .0=0
do đó mệnh đề này sai
d,$\displaystyle  \begin{array}{{>{\displaystyle}l}}
G\ là\ trọng\ tâm\ tam\ giác\ ADM\ nen\ ta\ có\ tỉ\ số\ 2:1\\
\overrightarrow{CG} =\frac{1}{3} .(\overrightarrow{CA} \ +\overrightarrow{CD} \ +\overrightarrow{CM})\\
\overrightarrow{CA} +\overrightarrow{DM} \ =\overrightarrow{CA} \ +(\overrightarrow{DA} -\overrightarrow{AM}) =\overrightarrow{CA} \ +(\overrightarrow{DA} \ -\frac{1}{2}\overrightarrow{AB)}
\end{array}$
⟹$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\overrightarrow{CG} .(\overrightarrow{CA} +\overrightarrow{DM}) =\frac{1}{3}(\overrightarrow{CA} +\overrightarrow{CD} +\overrightarrow{CM}) .\left(\overrightarrow{CA} +\overrightarrow{DA} \ \ -\frac{1}{2}\overrightarrow{AB}\right)( *)\\
\overrightarrow{CA} \ .\overrightarrow{CA} \ =a^{2} \ ,\overrightarrow{CD} \ .\overrightarrow{DA} \ =0,\overrightarrow{CM} .\overrightarrow{AB\ } \ =0\ ,\overrightarrow{CA} .\overrightarrow{DA} \ =0,\\
\overrightarrow{CD} \ .\overrightarrow{AB} \ =0\ ,\overrightarrow{CM} .\overrightarrow{DA\ } =0,\overrightarrow{CA} \ .\overrightarrow{AB} \ =0\ ,\overrightarrow{CM} .\overrightarrow{CD} \ =\frac{1}{2} a^{2}\\
\overrightarrow{CM\ } \ .\overrightarrow{DA} \ =\frac{-1}{2} a^{2}\\
thay\ vào\ ( *) \ ta\ có=\frac{1}{3} .\left( a^{2} +0+0+0+0+0+0+\frac{1}{2} a^{2} -\frac{1}{2} a^{2}\right) =\frac{1}{3} a^{2}\\
\ d\ sai\\
\end{array}$
$ $