When approaching this problem, i tried to use the trigonometric form of Ceva's theorem, and in fact, if you define the isogonal conjugate so that $P'A,P'C$ and $P'B$ are reflections of $PA,PC,PB$ respectively, then you can show that way that those lines are concurrent, for example this way:
But the thing is that when trying to solve it the other way around, when you have the concurrency and two of the angle relations, I tried reversing the previous proof, but if you apply Ceva to the triangle, and you use the two angle relations you have in those, you get,finally: $$ \frac{\sin\angle ACP}{\sin\angle PCB} \times \frac{\sin\angle ACP'}{\sin\angle P'CB} = 1$$
which is not enough to justify $$\angle ACP = \angle P'CB$$
Can someone explain me how to use this to get to the final result? I understand that it's basically the same question using the three angle conditions and then proving they concur, but i was just wondering if this also is posible.
Since$$\angle ACP+\angle PCB=\angle ACP^\prime+\angle P^\prime CB=\angle ACB,$$we just need to use the fact that$$\frac{\sin(\theta-x)}{\sin x}=\sin\theta\cot x-\cos\theta$$decreases as $x$ increases from $0$ to $\theta:=\angle ACB$. Since your sine ratios are reciprocals of each other rather than equal, it follows that$$\angle ACP=\theta-\angle ACP^\prime=\angle P^\prime CB,$$as required.