Let's start by drawing a figure, taking care to highlight the triangle $O O_{1} O_{2}$, since $P$ is the center of the circumcircle of this triangle. We also note $\omega_{1}, \omega_{2}$, and $\omega^{\prime}$ as the circumcircles of $A D G$, $A E F$, and $O O_{1} O_{2}$, respectively.
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We immediately notice that the points $A, O_{1}$, and $O_{2}$ seem to be collinear, so we quickly set out to prove this. Indeed, the lines $(D G)$ and $(E F)$ are parallel to each other. Thales' theorem then indicates that there is a homothety $h$, centered at $A$, which maps the triangle $A D G$ to the triangle $A F E$. We deduce that $O_{2}=h\left(O_{1}\right)$, and therefore that the points $A$, $O_{1}$, and $O_{2}$ are indeed collinear.
To prove that $(O P)$ is parallel to $\ell_{B}$ and $\ell_{C}$, we will now calculate the angle between the lines $\left(O P, \ell_{B}\right)$. To do this, we start by getting rid of the point $P$ by invoking the tangent at $O$ to $\omega^{\prime}$: if we denote this tangent by $t$, then

$$\begin{aligned} \left(O P, \ell_{B}\right) & =(O P, t)+\left(t, O O_{2}\right)+\left(O O_{2}, \ell_{B}\right) \\ & =90^{\circ}+\left(O_{1} O, O_{1} O_{2}\right)+\left(O O_{2}, \ell_{B}\right) \\ & =90^{\circ}+\left(O_{1} O, O_{1} A\right)+\left(O O_{2}, \ell_{B}\right) . \end{aligned}$$

Since $(A D)$ is the radical axis of the circles $\omega_{1}$ and $\omega^{\prime}$, the lines $(A D)$ and $\left(O O_{1}\right)$ are perpendicular to each other. Similarly, $(A E)$ and $\left(O O_{2}\right)$ are perpendicular. We deduce that

$$\begin{aligned} \left(O P, \ell_{B}\right) & =90^{\circ}+\left(O_{1} O, O_{1} A\right)+\left(O O_{2}, \ell_{B}\right) \\ & =90^{\circ}+\left(O_{1} O, A D\right)+\left(A D, O_{1} A\right)+\left(O O_{2}, A E\right)+\left(A E, \ell_{B}\right) \\ & =90^{\circ}+90^{\circ}+\left(A D, O_{1} A\right)+90^{\circ}+\left(A E, \ell_{B}\right) \\ & =90^{\circ}+\left(A D, O_{1} A\right)+\left(A E, \ell_{B}\right) \end{aligned}$$

By doing so, we have already eliminated the points $P, O$, and $O_{2}$, suggesting that we are on the right track. Our next victim will be the point $O_{1}$. Indeed, if we denote by $A_{1}$ the symmetric point of $A$ with respect to $O_{1}$, then $\left[A A_{1}\right]$ is a diameter of $\omega_{1}$, so

$$\left(A D, O_{1} A\right)=\left(A D, A A_{1}\right)=\left(A D, D A_{1}\right)+\left(D A_{1}, A A_{1}\right)=90^{\circ}+(D G, A G)=90^{\circ}+(B D, A E)$$

But then we have already won, since

$$\left(O P, \ell_{B}\right)=90^{\circ}+\left(A D, O_{1} A\right)+\left(A E, \ell_{B}\right)=90^{\circ}+90^{\circ}+(B D, A E)+\left(A E, \ell_{B}\right)=\left(B D, \ell_{B}\right),$$

which concludes the proof.
$\underline{\text { Alternative Solution } n^{\circ} 1}$ We reuse the names of the points and circles introduced in the previous solution.
Since the lines $\ell_{B}$ and $\ell_{C}$ are parallel, and since $A, B, C, D$, and $E$ are concyclic, we know that $(G A, G D)=(E A, E F)$ and, similarly, $(F E, F A)=(D G, D A)$. The triangles $A E F$ and $A G D$ are therefore similar. Consequently, the triangles $F O_{2} A$ and $D O_{1} A$ are also similar, so $\left(A F, A O_{2}\right)=\left(A D, A O_{1}\right)$, and therefore the points $O_{1}$, $A$, and $O_{2}$ are collinear.
Next, since the line $(A D)$ is the radical axis of the circles $\omega$ and $\omega_{1}$, it is perpendicular to $\left(O O_{1}\right)$. We deduce, from the theorem of the angle at the center, and by denoting $A_{1}$ the symmetric point of $A$ with respect to $O_{1}$, that

$$\begin{aligned} \left(O_{1} O_{2}, O_{1} O\right) & =\left(A A_{1}, A D\right)+\left(A D, O O_{1}\right)=\left(A A_{1}, D A_{1}\right)+\left(D A_{1}, A D\right)+90^{\circ} \\ & =(A G, D G)+90^{\circ}+90^{\circ}=(G A, G D) . \end{aligned}$$

We show similarly that $\left(O_{2} O, O_{2} O_{1}\right)=(F E, F A)=(D G, D A)$. The triangles $O O_{1} O_{2}$ and $A G D$ are therefore indirectly similar.
But then the triangles $O O_{1} P$ and $A G O_{1}$ are also indirectly similar. We conclude that

$$\left(O P, O_{1} O_{2}\right)=\left(G D, A O_{1}\right)=\left(G D, O_{1} O_{2}\right),$$

which means that $(O P)$ is parallel to the line $(G D)=\ell_{B}$, and therefore to $\ell_{C}$ as well.

Remark: If one decides to use directed angles, it is important to never divide angles by two, for example, to use the theorem of the angle at the center or when encountering an isosceles triangle or an angle bisector. Indeed, here is an example of a horror that could result if one does not take this precaution:
«Since the line $\left(O O_{1}\right)$ is the angle bisector of $\widehat{A O_{1} D}$, we know that $\left(O_{1} O, O_{1} A\right)=\left(O_{1} D, O_{1} A\right) / 2$. Since $A O_{1} D$ is isosceles at $O_{1}$, we deduce that

$$\left(A D, A O_{1}\right)=\left(\left(D O_{1}, D A\right)+\left(A D, A O_{1}\right)\right) / 2=\left(D O_{1}, A O_{1}\right) / 2=\left(O_{1} O, O_{1} A\right)$$

But this equality is of course completely false, since we actually have $\left(A D, A O_{1}\right)=\left(O_{1} O, O_{1} A\right)+90^{\circ}$. Oops!»
The reason why the above reasoning is false is that directed angles are angles modulo $180^{\circ}$. Consequently, if we divide a directed angle by two, we get a relation that is only valid modulo $90^{\circ}$. We will therefore avoid, at all costs, if we decide to use directed angles, dividing angles by two.

Comment from the graders: The problem was solved by a small number of people. However, many students managed to make significant progress in the problem, trying to prove what they could conjecture from their figure, which is an excellent approach. Simple observations, such as the fact that the lines $\left(O O_{1}\right)$ and $(A D)$ are perpendicular, were actually valuable for solving the problem and were thus rewarded. Do not hesitate to write down all your ideas, even seemingly trivial observations.