Using the distance formula,
[tex]AD=\sqrt{(-1-3)^{2}+(-3-(-4))^{2}}=\sqrt{17}\\\\BC=\sqrt{(5-9)^{2}+(1-0)^{2}}=\sqrt{17} \\\\\therefore \overline{AD} \cong \overline{BC}[/tex]
[tex]AB=\sqrt{(-1-5)^{2}+(-3-1)^{2}}=\sqrt{52}=2\sqrt{13}\\\\CD=\sqrt{(9-3)^{2}+(0-(-4))^{2}}=\sqrt{52}=2\sqrt{13}\\\\\therefore \overline{AB} \cong \overline{CD}[/tex]
Since ABCD has two pairs of opposite congruent sides, it is a parallelogram.
