Solution:

(a) To prove (a) it suffices to show that triangle $BDF$ is isosceles on base $BD$. Since $AF$ is the bisector of angle $\widehat{BAD}$, we have $\widehat{FAD} = \widehat{FAB}$. Moreover, $\widehat{FBD} = \widehat{FAD}$ since they subtend the same arc of the circle, and, for the same reason, $\widehat{FDB} = \widehat{FAB}$. Hence triangle $BDF$ is isosceles with base $BD$, since the two angles adjacent to the base $\widehat{FBD}$ and $\widehat{FDB}$ are equal.

(b) Let us now turn to point (b). We know that $\widehat{BAE} = \widehat{AED}$, as they are alternate interior angles of the two parallel segments $AB$ and $CD$ cut by the transversal $AE$. Moreover, since $\widehat{BAD}$ is right and $AE$ is its bisector, $\widehat{DAE} = \widehat{AED} = 45^\circ$. Since the quadrilateral $ADHF$ is inscribed in a circle, $\widehat{DHF} = 180^\circ - \widehat{DAF}$ and the angles $\widehat{DEA}$ and $\widehat{DEF}$ are supplementary, so $\widehat{DEF} = \widehat{DHF}$.

Now observe that, by construction, $EH$ is perpendicular to $DF$, so the two triangles $DEF$ and $DHF$ are symmetric with respect to $DF$ and therefore in particular the triangles $DEG$ and $DHG$ are congruent.

(c) Finally, regarding point (c), we begin by observing that from the previous point we have
$$\widehat{CDF} = \widehat{EDG} = \widehat{GDH} = \widehat{FDH} \text{;}$$
moreover, the angles $\widehat{FDH}$ and $\widehat{HCF}$ subtend the same arc and are therefore equal, as are the angles $\widehat{FHC}$ and $\widehat{CDF}$. We deduce that the angles $\widehat{FHC}$ and $\widehat{HCF}$ are equal, so triangle $FHC$ is isosceles with base $HC$ and the segments $FC$ and $FH$ have the same length.