a) Let $(A'), (B'), (C')$ respectively represent the circle passing through point $C$ and touching the line $AB$ at point $A$, the circle passing through point $B$ and touching the line $BC$ at point $B$, and the circle passing through point $C$ and touching the line $CA$ at point $C$.

Let $K$ be the second intersection point of two circles $(A')$ and $(B')$. We have
$$(AK, AB) \equiv (BK, BC) \equiv (CK, CA) \pmod{\pi}.$$
Therefore, point $K$ also belongs to circle $(C')$. Now, denote by $D, E, F$ respectively the foot of the perpendicular drawn from point $K$ to lines $BC, CA$ and $AB$. According to Erdos inequality, we have
$$KA + KB + KC \geq 2(KD + KE + KF).$$

Let $\angle KBC = \angle KAB = \angle KCA = \omega$. Because
$$\sin \omega = \frac{KD}{KB} = \frac{KE}{KC} = \frac{KF}{KA} = \frac{KD + KE + KF}{KB + KC + KA} \le \frac{1}{2}$$
so $\omega \le 30^\circ$.

The triangles $KAA'$, $KBB'$, $KCC'$ are isosceles triangles at $A'$, $B'$, $C'$ with vertex angle equal to $2\omega \le 60^\circ$.

Put
$$(\overrightarrow{KA}, \overrightarrow{KA'}) \equiv (\overrightarrow{KB}, \overrightarrow{KB'}) \equiv (\overrightarrow{KC}, \overrightarrow{KC'}) \equiv \phi \pmod{2\pi}$$
$$\text{and } k = \frac{KA'}{KA} = \frac{KB'}{KB} = \frac{KC'}{KC} \ge 1.$$
We denote by $f$ the rotational homothety with center $K$, angle $\phi$ and coefficient $k$.

Since $A'$, $B'$, $C'$ are images of $A$, $B$, $C$ by $f$ respectively, $\triangle A'B'C' \sim \triangle ABC$. We deduce that $\frac{S(A'B'C')}{S(ABC)} = k^2 \ge 1$, or $S(A'B'C') \ge S(ABC)$. The equality occurs if and only if $ABC$ is an equilateral triangle.

b) Since $A'$, $C'$ are images of $A$, $C$ through $f$ and $C'B' \perp BK$, $C'O \perp BC$ respectively, we get
$$(C'K, C'A') \equiv (CK, CA) \equiv (BK, BC) \equiv (C'B', C'O) \pmod{\pi}.$$
Therefore $C'O$ and $C'K$ are isogonal in angle $A'C'B'$. By similar argument, we have $O$ and $K$ are isogonal conjugate points in triangle $A'B'C'$. So $KX' \perp A'B'$, $KY' \perp B'C'$ and $KZ' \perp C'A'$. We also have $AK \perp A'B'$, $BK \perp B'C'$ and $CK \perp C'A'$ so we deduce that three lines $AX'$, $BY'$ and $CZ'$ concur at $K$.