We know that diagonals bisect each other perpendicularly, so we can draw the following picture:
Then, from the right triangle FED,
we have that
[tex]x^2+16^2=28^2[/tex]
which gives
[tex]x^2=784-256[/tex]
then
[tex]\begin{gathered} x^2=528 \\ x=\sqrt[]{528} \\ x=\sqrt[]{16\cdot33} \\ x=4\sqrt[]{33} \end{gathered}[/tex]
Therefore,
[tex]EF=x=4\sqrt[]{33}[/tex]
Since EC is twice EF, then
[tex]EC=8\sqrt[]{33}[/tex]
Finally, we know that all sides are equal, this means that
[tex]BC=ED=28=CD=BE[/tex]
In summary, the answers are:
[tex]\begin{gathered} CD=28 \\ FD=\frac{32}{2}=16 \\ EF=4\sqrt[]{33} \\ EC=8\sqrt[]{33} \end{gathered}[/tex]
