From the image given, the data shows that
[tex]\begin{gathered} <1\cong<3 \\ \text{and} \\ AD\cong BC \end{gathered}[/tex]
We can observe that
[tex]\begin{gathered} \Delta BDA\cong\Delta DBC\text{ (SAS)} \\ \text{ Reasons:} \\ AD\cong BC\Rightarrow side \\ \measuredangle1\cong\measuredangle3\Rightarrow\text{angle} \\ BD\cong DB(common\text{ sides or reflexive)}\Rightarrow\text{side} \end{gathered}[/tex]
Thus, from the above we can say that;
[tex]\begin{gathered} AB\cong DC\text{ (corresponding parts of congruent triangles are congruent)} \\ \text{Therefore, } \\ \measuredangle2\cong\measuredangle4 \end{gathered}[/tex]
Hence
Yes, the given information is enough to prove that ABCD is a parallelogram.
