BMO 2017

We have $(e)\perp BC$ and $I_{a}D\perp BC$, so $(e)\parallel I_{a}D$. Let $T$, $S$ be the midpoints of the segments $H{I}_a$, $HD$ respectively and $Y$ the point of intersection of the lines $HD$, $EZ$. Then, $TS\parallel I_{a}D$, $TS\perp BC$ and $SY\perp EZ$. The Euler circle $(\omega )$ of the triangle $EDZ$ passes through the points $N$, $Y$, $S$. Therefore, the segment $SN$ is a diameter of the circle $(\omega )$. Thus, the center of $(\omega )$, let $T^{\prime }$, is the midpoint of the segment $SN$.

On the other hand, we know that the center of Euler circle $(\omega )$ is the midpoint $T$ of $H{I}_a$. So $T=T^{\prime }$. Therefore, the line $(e)$ passes through the points $T$, $S$. Therefore, we get that the quadrilateral $HS{I}_{a}N$ is parallelogram and its diagonals meet each other at the point $T$. We consider the inversion $I(I_a,I_a{Z}^2)$. As $I_a{Z}^2=I_{a}A\cdot I_{a}N$ we have $I(N)=A$. Similarly, if $M_1$, $M_2$ the midpoints of the segments $DE$, $DZ$ respectively, we get, $I(M_1)=B$ and $I(M_2)=C$. Therefore, the circumcircle $(C)$ of the triangle $ABC$ is the image of the circle $(\omega )$ under the inversion $I$ and the points of the intersection of the circles and $(\omega )$ are invariant under this inversion. But it is well known that the circle of inversion passes through the points of the intersection of the circles $(C)$ and $(\omega )$. Thus, the Euler circle $(\omega )$ passes through the points $I$, $L$. Also, we consider the inversion $J(H,r^2)$ with $$r^2=HX\cdot HZ=HD\cdot HY=HW\cdot HE$$ where $X$, $W$, $Y$ the traces of the altitudes of the triangle $EDZ$ on its sides. Then, $J(Z)=X$, $J(D)=Y$ and $J(E)=W$. Therefore, the circumcircle $(C_1)$ of the triangle $ABC$ is the image of the circle $(\omega )$ under the inversion $J$. Thus, the circle of inversion $J$ passes through the points $I$, $L$. We conclude that $HI=HL$ and $H{I}_a\perp IL$ and since $(\delta )\parallel IL$, we have $H{I}_a\perp (\delta )$. If $R$ is the point of intersection of the lines $(\delta )$, $HL$, we get that quadrilateral $HR{I}_{a}G$ is parallelogram and its diagonals meet each other at the point $T$. So, the perpendicular line $(e)$ through the point $N$ to the line $BC$ and the parallel line $(\delta )$ through the point $G$ to the line $IL$, meet each other on the line $H{I}_a$.