Construction: Join ED.
The corresponding diagram is given below,
According to the given problem,
[tex]\begin{gathered} AE=BD \\ AE\parallel BD \end{gathered}[/tex]
Since a pair of opposite sides are parallel and equal, it can be claimed that quadrilateral ABDE is a parallelogram.
Then, as a property of any parallelogram, it can be argued that,
[tex]\begin{gathered} AB=DE \\ AB\parallel DE \end{gathered}[/tex]
Given that B is the mid-point of AC,
[tex]\begin{gathered} AB=BC \\ AB\parallel BC \end{gathered}[/tex]
Combining the above two results,
[tex]\begin{gathered} BC=DE \\ BC\parallel DE \end{gathered}[/tex]
It follows that ABCD also forms a parallelogram.
Again using the property that opposite sides of a parallelogram are equal and parallel. It can be claimed that,
[tex]\begin{gathered} EB=DC \\ EB\parallel DC \end{gathered}[/tex]
Hence proved that segment EB is parallel to segment DC,
[tex]\vec{EB}=\vec{DC}[/tex]
