Solution:

(a) Suppose all the six statements are true. Then $A B D E$, $B C E F$, $C D F A$ are isosceles trapeziums; if $K, L, M, P, Q, R$ are the mid-points of $A B$, $B C$, $C D$, $D E$, $E F$, $F A$ respectively, then we see that $K P \perp A B, E D$; $L Q \perp B C, E F$ and $M R \perp C D, F A$.

If $A D$, $B E$, $C F$ themselves concur at a point $O$, then $O A = O B = O C = O D = O E = O F$. ($O$ is on the perpendicular bisector of each of the sides.) Hence $A, B, C, D, E, F$ are concyclic and lie on a circle with centre $O$.

Otherwise these lines $A D$, $B E$, $C F$ form a triangle, say $X Y Z$. (See Fig.) Then $K X$, $M Y$, $Q Z$, when extended, become the internal angle bisectors of the triangle $X Y Z$ and hence concur at the incentre $O'$
of $X Y Z$. As earlier $O'$ lies on the perpendicular bisector of each of the sides. Hence $O' A = O' B = O' C = O' D = O' E = O' F$, giving the concyclicity of $A, B, C, D, E, F$.

(b) Suppose $(a_1)$, $(a_2)$, $(b_1)$, $(b_2)$ are true. Then we see that $A D = B E = C F$. Assume that $(c_1)$ is true. Then $C D$ is parallel to $A F$. It follows that triangles $Y C D$ and $Y F A$ are similar. This gives
$$\frac{F Y}{A Y} = \frac{Y C}{Y D} = \frac{F Y + Y C}{A Y + Y D} = \frac{F C}{A D} = 1$$
We obtain $F Y = A Y$ and $Y C = Y D$. This forces that triangles $C Y A$ and $D Y F$ are congruent. In particular $A C = D F$ so that $(c_2)$ is true. The conclusion follows from (a).

Now assume that $(c_2)$ is true; i.e., $A C = F D$. We have seen that $A D = B E = C F$. It follows that triangles $F D C$ and $A C D$ are congruent. In particular $\angle A D C = \angle F C D$. Similarly, we can show that $\angle C F A = \angle D A F$. We conclude that $C D$ is parallel to $A F$ giving $(c_1)$.