IMO Hk TST

Let $\ell_A$ meet $II'$ at $P$. We claim that $P$ is the intersection point of $\ell_A$, $\ell_B$, $\ell_C$. It suffices to show that $\frac{II'}{I'P}$ is a constant.
Let $DI'$ meet the incircle of $\triangle ABC$ again at $M$, and meet the line passing through $I$ and parallel to $\ell_A$ at $Q$. Let $r$ and $r'$ be the inradii of $\triangle ABC$ and $\triangle DEF$ respectively. By similar triangles, we obtain
$$\frac{II'}{I'P} = \frac{QI'}{I'D} = \frac{1}{I'D} \times \frac{MI' \times IA}{MA} = \frac{MI'}{I'D} \times \frac{IA}{MA}.$$
Since $I'$ is the incentre of $\triangle DEF$, we have $MI' = MF$. Thus, we have
$$\frac{MI'}{I'D} = MF \times \frac{1}{I'D} = \frac{d(M, EF)}{\sin \angle EFM} \times \frac{\sin \angle I'DF}{r'} = \frac{d(M, EF)}{r'}$$
since $\angle EFM = \frac{1}{2} \angle EDF = \angle I'DF$.

$$\frac{IA}{MA} = \frac{d(I, AF)}{d(M, AF)} = \frac{r}{d(M, EF)}$$
Combining these, we obtain
$$\frac{II'}{I'P} = \frac{MI'}{I'D} \times \frac{IA}{MA} = \frac{d(M, EF)}{r'} \times \frac{r}{d(M, EF)} = \frac{r}{r'}$$
This is a constant, and so $\ell_A$, $\ell_B$, $\ell_C$ are concurrent at $P$.