Since $ \triangle ELJ $ and $ \triangle IKF $ are homothetic, so $ IE, KL, JF $ are concurrent .

Let $X,Y$ are the intersection of $FJ,EI$ and $FJ,KL$, respectively We have: $\frac{FX}{XJ}=\frac{FI}{JE};\frac{FY}{YJ}=\frac{FK}{JL}$ It's easy to see that $\triangle FKI$ is similar to $\triangle JLE$ Therefore $X$ coincide $Y$ and we have qed Thing's I see places of $E,F$ are not important if $E,F$ are the arbitrary point on $AC,AB$, respectively, $IE,JF,KL$ are also concurrent Hoa nhài

Since $IK \cap EL= P_{\infty 1}$, $KF \cap LJ = P_{\infty 2}$ and $FI \cap JE = P_{\infty 3}$, these three points are collinear (they all lie on the line at infinity). Now by Desargues' theorem, $IE, KL, FJ$ are concurrent, as desired.