Since $\angle ABC + \angle CDA < 180^\circ$, point $A$ lies inside the circumcircle $\omega$ of triangle $BCD$. Denote by $C'$, $D'$ the second intersection of $\omega$ with rays $CA$, $DA$, respectively (Fig. 3). We angle chase:
$$\angle D'C'C = \angle D'DC = \angle ADC = \angle ACB.$$
Hence $BCC'D'$ is an isosceles trapezoid. Moreover, since $\angle C'AD' = \angle CAD = \angle BAC$, triangles $ABC$ and $AD'C'$ are similar by AA and in fact due to $BC = C'D'$ they are congruent. Point $A$ is thus the midpoint of the chord $CC'$ and $\angle OAC = 90^\circ$ follows.

Fig. 3

Let's frame the figure with respect to triangle $ABD$. Then $AC$ is the A-angle bisector. The Inscribed angle theorem states that the (reflex) angle $BOD$ is twice the (convex) angle $BCD$, hence for the size of the convex angle $BOD$ we get
$$\angle BOD = 360^\circ - 2 \cdot \angle DCB = 360^\circ - \angle ADC - \angle BCD - \angle ABC = \angle BAD.$$

Fig. 4

Therefore $O$ lies on the arc $BAD$ of the circumcircle of triangle $ABD$. Since $OB = OD$, point $O$ is the midpoint of that arc and thus it lies on the external A-angle bisector which is perpendicular to the A-angle bisector.