Prove that three circumferences in a triangle intersect in a point We've been given this exercise by our professor as an optional one, but I don't know the correct way to tackle it.

Let $T=\triangle\{A,B,C\}$ be a triangle. Consider three points $A', B', C'$ so that $A'\in [B,C]-\{B,C\}$, $B'\in [A,C]-\{A,C\}$ and $C'\in [A,B]-\{A,B\}$.
Now let $C_A$ be the circumference through $A, B', C'$, $\ C_B$ be the circumference through $B, A', C'$ and $C_C$ be the circumference through $C, A', B'$.
We will also assume that the intersection of every two circumferences from $\{C_A, C_B, C_C\}$ has two different points.
Prove that $C_A, C_B$ and $C_C$ intersect in a single point $P$.

I've drawn this and it makes sense obviously, but I'm struggling to find a way to prove this formally.

Thanks for your time.
 A: Points $A,A',B,B',C'$ are given. We do not talk about $C$ as of now. The supplements are shown  marked as it is known opposite angles in cyclic quadrilateral sum up to $180^0$.
The locus circle runs through three points $(A',B', M)$ subtending  in it angle $ (180^0-A-B)$ between extensions of sides $(AB',A'B).$
In other words the deficit angle from $360^0$ is $ 360^0-[(180^0-A)+(180^0-B)]= A+B$
Emanating from the intersection point $M$ draw the three green lines like a star as shown. Produce $AB',A'B$ to intersect at an unknown point $X$.
Since we are drawing a circle through $(A',B', X)$ to make a cyclic quadrilateral the sum of angles at $M$ and $X$ should be $180^0$ again ;
So the angle at $X = 180^0-A-B $ must be same as $C$, validating the starting/in-built  assumption of concurrency at Miguel point $M$.

A: Let the circumcircles of $\triangle AC'B'$ and $\triangle CA'B'$ intrsect at point $P$. Proving points $B$, $A'$, $C'$ and $P$ concyclic is what we aim to do.
$\angle A'PB'=180-\angle C$ and $\angle C'PB'=180-\angle A$ using the fact that opposite angles of a cyclic quadrilateral are supplementary.
Hence,  $\angle C'PA'=360-\angle A'PB'-\angle C'PB'=180-\angle B$ and thereafter $P$ lies on the circumcircle of $\triangle BA'C'$.
