Simple&Quick Angle Chasing.....

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From $\measuredangle AOC = \measuredangle XDC$ not implies that $AO || XD$ , it needs $D-O-C$ to be collinier, but you trying to prove that.

Inversion at $A$ with radius $AB$ and we're done.

Say, $\angle OAB=\angle OBA=\theta\implies\angle ACO=\angle OCB\implies\angle DAB=2\theta\implies\angle DAO=\angle ADO=\theta$. So, $\angle AOD=180^\circ-2\theta$ and as $AB=AC\implies\angle ABC=\angle AOC=2\theta\implies\angle AOD+\angle AOC=180^\circ\implies C-O-D$ are collinear.

This is what I submitted to someone on earth a long time back

Attachments:
Solution 2.pdf (234kb)

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Attachments:
Solution 2.pdf (234kb)

Solution

Attachments:

Please check my solution. Let $O_{2}C$ intersects $w_{2}$ at $E$. Then $\angle BO_{2}D=2\angle BEO_{2}=2a$. Since $\triangle BEO_{2}$ is isosceles, $EO_{2}B=180-2a$. This implies collinear because $C,E,O_{2}$ is collinear.

Redefine $C=\overline{OD}\cap\omega_3.$ Notice $\triangle AOD\cong\triangle AOB$ so $$\angle OCA=\angle ADC=\angle OBA$$and $C$ lies on $\omega_1.$ $\square$

Not the simplest solution in this thread, but a new (way too detailed) solution Let the centre of $\omega_1 = P$ We want to prove that $$\angle COB+\angle BOD=180^{\circ} \iff 2 \cdot \angle CAB + \angle BAD = 180^{\circ} \iff 2 \cdot \angle CAB + 2 \cdot \angle BAP = 180^{\circ} \iff \angle CAB+\angle BAP = 90^{\circ} \iff AC \ \text{is tangent to} \ \omega_1$$This is immediate by inversion with centre $O$ and radius $OA \ \blacksquare$

Just take $D$ is $C$-excircle center and by angle chasing it have to lies on $\omega_3$