Let $A=(-a,0)$, $B=(b,d)$, $C=(c,e)$, and $D=(0,0)$. Let $p=-a=AD$, $q=\sqrt{b^2+d^2}=BD$, and $r=\sqrt{c^2+e^2}=CD$. We have $AE+BE=p+q$ and $BE+CE=q+r$, so $AE=(p+q-BE)$ and $CE=(q+r-BE)$. Therefore, $(q+r)AE^2+(p-r)BE^2-(p+q)CE^2$ is constant and describes the equation of line $DE$. Therefore, if $E=(x,y)$, this is $$(q+r)(x+a)^2+(p-r)(x-b)^2-(p+q)(x-c)^2+(q+r)y^2+(p-r)(y-d)^2-(p+q)(y-e)^2$$so the equation of $DE$ is $$x(aq+ar+br+cq+ac-ab)+y(ae-ad+dr+eq)=0.$$The equation of the ellipse with foci $B$ and $C$ passing through $D$ and $E$ is \begin{align*} BE&=q+r-CE\\ BE^2&=(q+r)^2+CE^2-2(q+r)CE\\ (BE^2-CE^2-(q+r)^2)^2&=4(q+r)^2CE^2\\ (BE^2-CE^2)^2-2(q+r)^2(BE^2+CE^2)+(q+r)^4&=0\\ (-2bx+2cx-2dy+2ey+b^2-c^2+d^2-e^2)^2-2(q+r)^2(2x^2-2(b+c)x+2y^2-2(d+e)y+b^2+c^2+d^2+e^2)+(q+r)^4&=0\\ 4x^2((c-b)^2-(q+r)^2)+4y^2((e-d)^2-(q+r)^2)+8xy(c-b)(e-d)+4x((c-b)(q^2-r^2)+(b+c)(q+r)^2)+4y((e-d)(q^2-r^2)+(d+e)(q+r)^2)&=0\\ (x(c-b)+y(e-d))^2-(q+r)^2(x^2+y^2)+2(q+r)(x(qc+rb)+y(qe+rd))&=0. \end{align*} Since $D$ lies on the angle bisector, $(a+b)e+(a+c)d=0$, so $a=-\frac{be+cd}{d+e}$. Then, let $m=aq+ar+br+cq+ac-ab$ and $n=ae-ad+dr+eq$. If $x=Cn$ and $y=-Cm$, then we get \begin{align*} C((n(c-b)-m(e-d))^2-(q+r)^2(m^2+n^2))&=-2(q+r)(n(qc+rb)-m(qe+rd))\\ C((r(cd-be)+q(cd-be)-a(q+r)(e-d))^2-(q+r)^2(m^2+n^2))&=-2a(q+r)((e-d)(qc+rb)-(q+r+c-b)(qe+rd))\\ C(q+r)^2((cd-be-a(e-d))^2-m^2-n^2)&=-2a(q+r)(q(be-cd)+r(be-cd)-(q+r)(qe+rd))\\ C((cd-be-a(e-d))^2-m^2-n^2)&=2a(-be+cd+qe+rd). \end{align*} Therefore, if $s=(cd-be-a(e-d))^2-m^2-n^2$ and $t=-be+cd+qe+rd$, then $C=\frac{2at}s$, so $x=\frac{2atn}s$ and $y=\frac{-2atm}s$. The slope of $AE$ is $\frac{-2atm}{2atn+as}=\frac{-2tm}{2tn+s}$. Then, we have \begin{align*} 2tn+s&=2tn+(cd-be-a(e-d))^2-m^2-n^2\\ &=2tn+(t-n)^2-m^2-n^2\\ &=t^2-m^2. \end{align*} Therefore, the slope of $AE$ is $\frac{-2tm}{t^2-m^2}$. Now, we have \begin{align*} m&=q(a+c)+r(a+b)+a(c-b)\\ &=q\frac{ce-be}{d+e}+r\frac{bd-cd}{d+e}+\frac{(b-c)(be+cd)}{d+e}\\ &=\frac{b-c}{d+e}(-qe+rd+be+cd). \end{align*}Then, if $t'=be-cd+qe+rd$ and $m'=\frac{c-b}{d+e}(qe-rd+be+cd)$, the slope of $AF$ is $\frac{-2t'm'}{t'^2-m'^2}$, so we need to show $$\frac{-2tm}{t^2-m^2}=-\frac{-2t'm'}{t'^2-m'^2}$$or $$\frac tm-\frac mt=-\frac{t'}{m'}+\frac{m'}{t'}.$$We claim $\frac tm=-\frac{t'}{m'}$, which implies the equality. This is equivalent to \begin{align*} (-be+cd+qe+rd)\frac{c-b}{d+e}(qe-rd+be+cd)&=-(be-cd+qe+rd)\frac{b-c}{d+e}(-qe+rd+be+cd)\\ (-be+cd+qe+rd)(qe-rd+be+cd)&=(be-cd+qe+rd)(-qe+rd+be+cd)\\ (qe+cd)^2-(be-rd)^2&=(rd+be)^2-(qe-cd)^2\\ (rd+be)^2+(rd-be)^2&=(qe+cd)^2+(qe-cd)^2\\ r^2d^2+b^2e^2&=q^2e^2+c^2d^2\\ d^2(c^2+e^2)+b^2e^2&=(b^2+d^2)e^2+c^2d^2. \end{align*} This is true, which implies that the slopes of $AE$ and $AF$ have quotient $-1$. Therefore, $AE$ and $AF$ are reflections over $AD$, so $AD$ bisects $EAF$.

how long did that take @above

Rename $D$ as $I$ and let $D$ be the intersection of $y$ and $z$ other than $I$. Let $\gamma_A=(A,AI)$. Define $\gamma_B$ and $\gamma_C$ similarly. Claim. $D$ is the center of the circle $(D)$ which is internally tangent to $\gamma_A$ and externally tangent to $\gamma_B$ and $\gamma_C$. Proof. It suffices to show that $AI-AD=BD-BI$ but is given since $\{I,D\}\in z$. Define $(E)$ and $(F)$ similarly. Let $P,Q,R$ be reflections of $I$ over $BC$, $CA$, $AB$. Now invert the diagram in $I$. The circles $\gamma_A$, $\gamma_B$, and $\gamma_C$ form a triangle $P'Q'R'$ and the circles $(D)'$, $(E)'$, and $(F)'$ are the excircles of this triangle. Let $\gamma_A$ touch $(E)$ and $(F)$ at $Y$ and $Z$. then clearly $A$, $Y$, $E$ are colinear and so are $A$, $Z$, $F$. Under inversion, $Y$ and $Z$ map to the tangency points $Q'R'$ with the excircles. Since $I$ is on the $A$-angle bisector, $AI$ is the perpendicular bisector of $QR$, and thus is the perpendicular bisector of $Q'R'$. Thus $IY'=IZ'$, so $IY=IZ$. Hence $\angle YAI=\angle ZAI$. Extra: If $I$ is incenter, then $AD,BE,CF$ concur.

bachkieu wrote: how long did that take @above around 2 hours