Answer
First of all, we know that angle DEB is congruent to angle AEC by the vertical angles theorem. In addition, since the radius of a circle is always perpendicular to the tangent line at that point, it follows that BDE is congruent to ACE. Thus, by AA, the triangles are similar, allows us to set up this proportion:
$\frac{BD}{AC} = \frac{ED}{CE} \\ BDCE = ACED$
