#### Answer

EB = 6
FC = 7
AD = 3

#### Work Step by Step

We know that AD and AF are equal, so we know:
$ 10 - FC = 9 - BD$
We also know that EC equals FC, so:
$ 10 - AF = 13 - BE$
Finally, we know that BD equals BE.
From these, it follows that EB must be 3 more than AF. Thus, we call AF the variable y to obtain:
$EB = 3 + y$
$ BD = EB = 9 - y$
We solve for y:
$ 3+y = 9 -y \\ y = 3$
Thus:
$EB = 3 + 3 = 6$
$ AD = AF = y = 3$
$FC = 10 - y = 10 -3 = 7$