Concurrency of three lines in a multiple tangent circles configuration

Let $ABC$ be a triangle, $\mathcal{C}$ its circumscribed circle and $\mathcal{I}$ its inscribed circle. We construct a circle that is tangent on the interior to $\mathcal{C}$ in the point $A$ and is simultaneously exterior tangent to $\mathcal{I}$ in the point $A'$. Analogously, we obtain the points $B'$ and $C'$. Prove that the lines $AA', BB', CC'$ are concurrent.

I have tried using Ceva's theorem in the triangle $ABC$, by computing the sines of the angles $\angle BAA'$ and $\angle CAA'$, which could have helped to determine the ratio of segments determined by $AA'$ on the side $BC$.

The problem appeared in a set meant to be solved using geometric transformations, but in this case I don't see exactly how.

Inspired by the drawing, I have managed to prove that the lines $AA',BB',CC'$ pass through the same point on the line $OI$, therefore are concurrent.

Denote $CC'\cap OI=\{C''\}$ and with $r_C$ the radius of the circle tangent to $\mathcal{C}$ in $C$ and to $\mathcal{I}$ in $C'$. The following Lemma is useful:

Lemma: In a triangle $ABC$ with $D\in BC$ we have $\frac{BD}{CD}=\frac{AB}{AC}\cdot \frac{\sin \angle BAD}{\sin\angle CAD}$.

Aplying the Lemma in $\triangle ICF$ we obtain $\frac{r}{r_C}=\frac{CI}{r_C}\cdot\frac{\sin\angle ICC'}{\sin \angle FCC'}$. Applying the Lemma in $\triangle ICO$ we obtain $$\frac{IC''}{OC''}=\frac{CI}{R}\cdot\frac{\sin\angle ICC'}{\sin \angle FCC'}=\frac{r}{R},$$ where $r,R$ are the radii of the inscribed and circumscribed circles to the triangle $ABC$. This finishes the proof.