## 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 $M R \perp C D, F A$.

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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 $O^{\prime}$ of $X Y Z$. As earlier $O^{\prime}$ lies on the perpendicular bisector of each of the sides. Hence $O^{\prime} A=O^{\prime} B$ $=O^{\prime} C=O^{\prime} D=O^{\prime} E=O^{\prime} F$, giving the concyclicity of $A, B, C, D, E, F$.
(b) Suppose $\left(\mathrm{a}_{1}\right),\left(\mathrm{a}_{2}\right),\left(\mathrm{b}_{1}\right),\left(\mathrm{b}_{2}\right)$ are true. Then we see that $A D=B E=$ $C F$. Assume that ( $\mathrm{c}_{1}$ ) 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 ( $\mathrm{c}_{2}$ ) is true. The conclusion follows from (a). Now assume that ( $\mathrm{c}_{2}$ ) 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 $\left(\mathrm{c}_{1}\right.$ ).