Japan Mathematical Olympiad Final Round

Let $X$ be the point of intersection of line $BC$ and the line going through point $D$ and parallel to line $AC$. Then, we see that $X$ lies on the line going through point $E$ and parallel to line $AB$, because we have $BX:XC=BD:DA=AE:EC$. Since line $IA$ (which is the same as $\ell$) is tangent to circle $\Gamma$, we have $\angle IAC=\angle ABC$. We also have $\angle ABC=\angle IXC$, since lines $EX$ and $AB$ are parallel. Consequently, we have $\angle IAC=\angle IXC$, from which it follows that 4 points $A,I,C,X$ are concyclic. By the theorem on powers of a point with respect to a circle, we then get $AE\cdot CE=IE\cdot XE$. By the same theorem, we also get $AE\cdot CE=FE\cdot GE$. Combining these two identities, we get $IE\cdot XE=FE\cdot GE$, which in turn implies that 4 points $I,X,F,G$ are concyclic. Similarly, we can obtain the fact that 4 points $H,X,F,G$ are concyclic, and therefore, 5 points $F,G,H,I,X$ are concyclic. Finally, from the fact that lines $AC$ and $HX$ are parallel, we get $\angle IHX=\angle IAC$, and since we also have $\angle IAC=IXC$ as we have seen earlier, we obtain the fact that $\angle IXC=\angle IHX$, and this implies that line $BC$ is tangent to the circumcircle of triangle $HIX$ at point $X$. Thus we have shown that points $F,G,H,I$ lie on a same circle and that circle is tangent to line $BC$.