Let $A$-Ex point($X_A$) be $EF \cap BC$ . It is well-known that $T$ is the $A$-Queue point of $\triangle ABC$. By Radical Axis Theorem on $\odot (ATBC),\odot (ATEF),\odot (BCEF)$, we get that $\overline{AT},\overline{EF},\overline{BC}$ concur; which gives that $A,T,X_A$ are collinear. Since $X_A$ is also on $XY$ , by Power of a point ,we are done.

Radical Axis kills this problem

Find cyclic quadrilaterals and use Radical Axis the XY intersects with all 3 radical Axises after this use power of a point And we are done.

Synthetic solution is easier but here is a short complex bash::: Set $(ABC)$ as the unit circle in the complex plane. $a'=\overline{a}$ so that $|a|=|b|=|c|=1$ and here is the equation of other points::: It's well known that::: $e=\frac{1}{2}(a+b+c-ac\overline{b})$ $f=\frac{1}{2}(a+b+c-ab\overline{c})$ If you do enough calculations and you have a bit patience you get::: $x=\frac{\frac{a\overline{b}}{2}(c'b-b'c)+\frac{1}{4}([\frac{ab+bc+ca-b^2}{abc}][a+b+c-b'ac]-[\frac{ab+bc+ca-c^2}{abc}][a+b+c-c'ab])}{\frac{1}{2}(c^2-b^2)}$ $y=\frac{\frac{a\overline{b}}{2}(-c'b+b'c)+\frac{1}{4}([\frac{ab+bc+ca-c^2}{abc}][a+b+c-c'ab]-[\frac{ab+bc+ca-b^2}{abc}][a+b+c-b'ac])}{\frac{1}{2}(-c^2+b^2)}$ Using that $ATEF$ is cyclic and getting an equation for $t$ then we can prove that $AXYT$ is cyclic::: $f(ATEF)=f(\overline{ATEF})$ $f(AXYT)=^?f(\overline{AXYT})$ where $f(ABCD)=\frac{c-a}{c-b}: \frac{d-a}{d-b}$

You can write a^-1 instead of a' same for b and c bc they lie on unit circle

wizixez wrote: Synthetic solution is easier but here is a short complex bash::: Set $(ABC)$ as the unit circle in the complex plane. $a'=\overline{a}$ so that $|a|=|b|=|c|=1$ and here is the equation of other points::: It's well known that::: $e=\frac{1}{2}(a+b+c-ac\overline{b})$ $f=\frac{1}{2}(a+b+c-ab\overline{c})$ If you do enough calculations and you have a bit patience you get::: $x=\frac{\frac{a\overline{b}}{2}(c'b-b'c)+\frac{1}{4}([\frac{ab+bc+ca-b^2}{abc}][a+b+c-b'ac]-[\frac{ab+bc+ca-c^2}{abc}][a+b+c-c'ab])}{\frac{1}{2}(c^2-b^2)}$ $y=\frac{\frac{a\overline{b}}{2}(-c'b+b'c)+\frac{1}{4}([\frac{ab+bc+ca-c^2}{abc}][a+b+c-c'ab]-[\frac{ab+bc+ca-b^2}{abc}][a+b+c-b'ac])}{\frac{1}{2}(-c^2+b^2)}$ Using that $ATEF$ is cyclic and getting an equation for $t$ then we can prove that $AXYT$ is cyclic::: $f(ATEF)=f(\overline{ATEF})$ $f(AXYT)=^?f(\overline{AXYT})$ where $f(ABCD)=\frac{c-a}{c-b}: \frac{d-a}{d-b}$ Bro... I respect your patience