Turkey — Team Selection Test

In the solution, we use directed lengths. Let $\ell$ intersect $EF$, $DF$, $DE$ at $K$, $L$, $M$ respectively and let $IX = x$, $IY = y$ and $IZ = z$; hence $IK = r^2/x$, $IL = r^2/y$ and $IM = r^2/z$ where $r$ is the radius of the incircle. Consequently we have
$$(1) \quad \frac{MX}{LX} \cdot \frac{KY}{MY} \cdot \frac{LZ}{KZ} = \frac{r^2/z - x}{r^2/y - x} \cdot \frac{r^2/x - y}{r^2/z - y} \cdot \frac{r^2/y - z}{r^2/x - z} = 1.$$
Now let the lines $DX$, $EY$, $FZ$ intersect $EF$, $DF$, $DE$ at $X'$, $Y'$, $Z'$ respectively; hence the lines $DE$, $DF$, $DX$, $DK$ intersect $EF$ at

$E$, $F$, $X'$, $K$ respectively and intersect $\ell$ at $M$, $L$, $X$, $K$ respectively, thus we have an equality of cross-ratios
$$(E, F; X', K) = (DE, DF; DX', DK) = (M, L; X, K), \quad \text{that is,}$$
$$\frac{EX'/FX'}{EK/FK} = \frac{MX/LX}{MK/LK}.$$
Multiplying the last equality with the analogous equalities
$$\frac{FY'/DY'}{FL/DL} = \frac{KY/MY}{KL/ML} \quad \text{and} \quad \frac{DZ'/EZ'}{DM/EM} = \frac{LZ/KZ}{LM/KM}$$
and rearranging, one finds
$$\frac{EX'}{FX'} \cdot \frac{FY'}{DY'} \cdot \frac{DZ'}{EZ'} = \left(\frac{EK}{FK} \cdot \frac{FL}{DL} \cdot \frac{DM}{EM}\right) \cdot \left(\frac{LK}{MK} \cdot \frac{ML}{KL} \cdot \frac{KM}{LM}\right) \cdot \left(\frac{MX}{LX} \cdot \frac{KY}{MY} \cdot \frac{LZ}{KZ}\right) = 1 \cdot (-1) \cdot 1 = -1,$$
where we used Menelaus' theorem in triangle $DEF$ with line $\ell$ and we used (1). Therefore, we conclude by Ceva's theorem in triangle $DEF$ that the lines $DX'$, $EY'$, $FZ'$ are concurrent. Done.

Let $D'$, $E'$, $F'$ denote the reflections of $D$, $E$, $F$ in the line $\ell$. Now $X$, $E$, $F$ are on the circle with diameter $[IA]$, hence $\angle XEF = \angle XIF$. On the other hand, $\angle XIF$ is the central angle of an arc of the incircle which measures half of the arc $\text{F'F}$, hence $\angle XIF = \angle F'EF$. As a result, $\angle XEF = \angle F'EF$ which implies that $X \in EF'$. Analogously, $X \in E'F$, $Y \in DF'$, $Y \in D'F$, $Z \in DE'$ and $Z \in D'E$. Now let $T$ denote the intersection of $EF$ and $E'F'$; by symmetry, we have $T \in \ell$. As a result, the points
$$F' = DY \cap XE, \quad E' = DZ \cap XF, \quad T = YZ \cap EF$$
are collinear, which implies that the triangles $DYZ$ and $XEF$ are perspective, hence the lines $DX$, $EY$, $FZ$ are concurrent (Desargues' theorem).