Let $ABC$ be a triangle with incenter $I$. Prove that the Euler lines of triangles $AIB, BIC, CIA,$ and $ABC$ are concurrent (called the Schiffler point of $ABC$).
Here is my approach :
Set the circumcircle of $ABC$ as the unit circle, hence we have $\overrightarrow{OH}=a+b+c$, where $O$ is the circumcenter and $H$ is the orthocenter of triangle $ABC$. Next, I attempted to find the Euler line of triangle $AIB$, but I will find the line connecting the circumcenter and the centroid of triangle $AIB$ instead (since the centroid also lies on the Euler line). Now let $z_{1}$ and $g_{1}$ denote the circumcenter and the centroid of triangle $AIB$ respectively. We then have $g_{1}=\frac{a+b+i}{3}$ and $$z_{1}=\frac{\begin{vmatrix} a&|a|^2&1\\ i&|i|^2&1\\ b&|b|^2&1 \end{vmatrix}}{\begin{vmatrix} a&\bar{a}&1 \\ i&\bar{i}&1 \\ b&\bar{b}&1 \end{vmatrix}}=\frac{|i|^2 - 1}{\bar{i}-\frac{a+b}{ab}+\frac{i}{ab}}$$ after some manipulations. But I need to find $\overrightarrow{Z_{1}G_{1}}$, which will be too ugly if I simply do $g_{1}-z_{1}$. The key point is that, I need to find some $k \in \mathbb{R}$ such that $k(a+b+c)$ lies on line $Z_{1}G_{1}$, and the same for the other two triangles. Can anyone find a better proof of this, or perhaps make the calculations simpler? Thanks for any help!
Let's set $(ABC)$ as the unit circle. Then, let's set $a=u^2,b=v^2,c=w^2$ for complex numbers $u,v,w$. We characterise the intersection of the Euler Lines of triangles $AIB, BIC, CIA$ as the intersection of the lines joining the orthocentre and centroid of each triangle.
Now by the Incentre-Excentre Lemma it follows that the circumcentre of $AIB$ say $O_1$ is the midpoint of $\overset{ \frown}{AB}$ not containig C.
Thus we've $o_1=-uv$. Then the centroid (say $G_1$), whose existence is an affine property is just the arithmetic mean of the vertices i.e. $g_1=(a+b+i)/3$. Now by the theorem on complex incentre we've $i=-(uv+vw+wu)$. Therefore we've $g_1=u^2+v^2-uv-vw-wu$.
Now we note that the as the orthocentre of $ABC$ is $u^2+v^2+w^2$ then the it's Euler line can be defined as all points $k(u^2+v^2+w^2)$ for all real numbers $k$
Now we find $k$ such that $k(u^2+v^2+w^2),O_1,G_1$ are collinear. Thus
$$\begin{align*} \frac{-uv-k(u^2+v^2+w^2)}{-uv-\frac13(u^2+v^2-uv-vw-wu)}&=\overline{\left(\frac{-uv-k(u^2+v^2+w^2)}{-uv-\frac13(u^2+v^2-uv-vw-wu)}\right)}\\ \frac{uv+k(u^2+v^2+w^2)}{u^2+v^2+2uv-vw-wu}&=\frac{\frac1{uv}+k\left(\frac1{u^2}+\frac1{v^2}+\frac1{w^2}\right)}{\frac1{u^2}+\frac1{v^2}+\frac2{uv}-\frac1{vw}-\frac1{wu}}\\ \frac uv+\frac vu+2-\frac uw-\frac vw+k(u^2+v^2+w^2)\left(\frac1{u^2}+\frac1{v^2}+\frac2{uv}-\frac1{vw}-\frac1{wu}\right)&=\frac uv+\frac vu+2-\frac wu-\frac wv+k\left(\frac1{u^2}+\frac1{v^2}+\frac1{w^2}\right)\left(u^2+v^2+2uv-vw-wu\right)\\ \end{align*}$$
Further computation gives
$$\begin{align*} k&=\frac{\frac uw+\frac vw-\frac wu-\frac wv}{(u^2+v^2+w^2)\left(\frac1{u^2}+\frac1{v^2}+\frac2{uv}-\frac1{vw}-\frac1{wu}\right)-\left(\frac1{u^2}+\frac1{v^2}+\frac1{w^2}\right)\left(u^2+v^2+2uv-vw-wu\right)}\\ &=\frac{\frac uw+\frac vw-\frac wu-\frac wv}{\frac{2w^2}{vu}-\frac{u^2}{vw}-\frac{v^2}{wu}+\frac{wv}{u^2}+\frac{w^2}{u^2}+\frac{wu}{v^2}+\frac{w^2}{v^2}-\frac{u^2}{w^2}-\frac{2uv}{w^2}-\frac{v^2}{w^2}}\\ &=\frac{uvw(u+v)(uv-w^2)}{(u+v)(uv-w^2)(uvw-u^2v-uv^2-v^2w-vw^2-w^2u-wu^2)}\\ &=\frac{uvw}{uvw-u^2v-uv^2-v^2w-vw^2-w^2u-wu^2},\\ \end{align*}$$
As it's symmetric in $u,v,w$ it follows that
$$\frac{uvw(u^2+v^2+w^2)}{uvw-u^2v-uv^2-v^2w-vw^2-w^2u-wu^2}$$
lies on all four Euler Lines i.e. of the triangles $ABC,AIB,BIC,CIA$ and we're done.