#### Answer

(a) $\triangle ABE \cong \triangle DFC$
(b) $\angle DAB$ and $\angle BCD$ are supplementary.

#### Work Step by Step

(a) Since $\angle ABD$ subtends the diameter of the circle, $m\angle ABD = 90^{\circ}$
Thus $m\angle ABE = 90^{\circ}$
It is given that $m\angle DFC = 90^{\circ}$
Thus $\angle DFC \cong \angle ABE$
Since $\angle ACD$ subtends the diameter of the circle, $\angle ACD = 90^{\circ}$
Thus $\angle ECD = 90^{\circ}$
Then $m\angle ECF + m\angle FCD = 90^{\circ}$
Since $m\angle FEC + m\angle ECF = 90^{\circ}$, then $m\angle FEC = m\angle FCD$
Also, $m\angle FEC = m\angle BEA$ since they are opposite angles.
Thus $m\angle FCD = m\angle BEA$, and so $\angle FCD \cong \angle BEA$
Since two angles of the triangles $\triangle ABE$ and $\triangle DFC$ are congruent, the third angle must also be congruent since the sum of three angles in a triangle is $180^{\circ}$
Therefore, $\triangle ABE \cong \triangle DFC$
(b) Since the quadrilateral $ABCD$ is a cyclic quadrilateral, the sum of opposite angles in the quadrilateral is $180^{\circ}$
Therefore, $m\angle DAB + m\angle BCD = 180^{\circ}$, so $\angle DAB$ and $\angle BCD$ are supplementary.