67th Czech and Slovak Mathematical Olympiad

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

Fig. 3

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Alternative solution.

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.