a. Xét D(x; y). Ta có: $\vec{AB}$ = (1; 3); $\vec{DC}$ = (5 - x; 5 - y)
Để ABCD là hình bình hành khi và chỉ khi $\vec{AB}$ = $\vec{DC}$
$\Leftrightarrow$ $\left\{\begin{matrix}5 - x = 1\\ 5 - y = 3\end{matrix}\right.$ $\Leftrightarrow$ $\left\{\begin{matrix} x = 4\\ y = 2\end{matrix}\right.$
Vậy D(4; 2)
b. Gọi M là giao điểm hai đường chéo của hình bình hành ABCD.
$\Rightarrow$ $\left\{\begin{matrix}x_{M}= \frac{x_{A} + x_{C}}{2}\\ y_{M} = \frac{y_{A}+y_{C}}{2}\end{matrix}\right.$ $\Leftrightarrow$ $\left\{\begin{matrix}x_{M}= \frac{2 + 5}{2}\\ y_{M} = \frac{2+5}{2}\end{matrix}\right.$ $\Leftrightarrow$ $\left\{\begin{matrix}x_{M}= \frac{7}{2}\\ y_{M} = \frac{7}{2}\end{matrix}\right.$
Vậy M($\frac{7}{2}$; $\frac{7}{2}$)
c. Ta có: $\vec{AC}$ = (3; 3), $\vec{BC}$ = (2; 0)
Suy ra: AB = |$\vec{AB}$| = $\sqrt{1^{2} + 3^{2}}$ = $\sqrt{10}$
AC = |$\vec{AC}$| = $\sqrt{3^{2} + 3^{2}}$ = $3\sqrt{2}$
BC = |$\vec{BC}$| = $\sqrt{2^{2} + 0^{2}}$ = 2
cosA = cos($\vec{AB}$,$\vec{AC}$) = $\frac{\vec{AB}.\vec{AC}}{AB.AC}$ = $\frac{1.3+3.3}{\sqrt{10}.3\sqrt{2}}$ = $\frac{2\sqrt{5}}{5}$ $\Rightarrow$ $\widehat{A}$ $\approx$ $26^{\circ}34'$
cosB = cos($\vec{BA}$,$\vec{BC}$) = $\frac{\vec{BA}.\vec{BC}}{BA.BC}$ = $\frac{(-1).2+(-3).0}{\sqrt{10}.2}$ = $\frac{-\sqrt{10}}{10}$ $\Rightarrow$ $\widehat{B}$ $\approx$ $108^{\circ}26'$
cosC = cos($\vec{CA}$,$\vec{CB}$) = $\frac{\vec{CA}.\vec{CB}}{CA.CB}$ = $\frac{(-3).(-2)+(-3).0}{3\sqrt{2}.2}$ = $\frac{\sqrt{2}}{2}$ $\Rightarrow$ $\widehat{C}$ = $45^{\circ}$