Lemma: if r, s are rays starting from a point O and A, B are points moving on r, s respectively, then AB passes through a fixed point P (strictly between those rays) if and only if there exist positive constants u, v, w such that u/a + v/b = w. Proof: If [AOB] denotes the area of triangle AOB, we have ab * 0.5 * sin(AOB) = [AOB] = [AOP] + [POB] = OP * (a * 0.5 * sin(AOP) + b * 0.5 * sin(BOP)). Divide both sides by 0.5 * ab to get the only if direction. For the if direction, note that we can easily find P by computing what sin(AOP) should be based on the previous equation. A corollary by inversion is: (AOB) passes through a fixed point P' if and only if there exist positive constants u, v, w such that u*a + v*b = w. Back to the main problem, note that CF * cos A + CE * cos B = CA * cos A + BC * cos B - AB/2 = constant > 0, so (CEF) indeed passes through a fixed point. Take D = A, B to easily conclude it must be the circumcenter of ABC.

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Almost the same configuration

Hey guys! The main idea is to complete the diagram by adding the Miquel point of the quadrilateral $AFEB$. Here is the solution: Let D´ be the reflection of D over the line $EF$. Since $\angle{ADF}$ = $\angle{FAD}$ = $\angle{A}$, and $\angle{EDB}$ = $\angle{B}$ (analogously), then $\angle{EDF}$ = $\angle{C}$. So, by the reflection: $\angle{ED'F}$ = $\angle{C}$. We can conclude that D' lies on $(CEF)$. Again, the reflection also gives us that $D'E$ = $DE$ = $EB$. Then, E is the center of $(BDD')$. By this, we have: $\angle{BD'D}$ = $\frac{1}{2}\angle{BED}$ = $90º$ - $\angle{B}$. Analogously: $\angle{CD'D}$ = $90º$ - $\angle{A}$. So: $\angle{BD'C}$ = $\angle{C}$ and D' must lie on $(ABC)$. (In fact, D' is the Miquel point of the quadrilateral $AFEB$). Noting that AD' is the radical axis of $(ABC)$ and $(DAD')$, we conclude that $OF$ is perpendicular to $AD'$. Analogously, $OE$ is perpendicular to $BD'$. Then: $\angle{EOF}$ = $180º$ - $\angle{AD'B}$ = $180º$ - $\angle{C}$ = $180º$ - $\angle{ED'F}$. As we want to show: O lies on $(CEF)$.

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Really good and nice problem! Mehhhhhhh Let $\angle A=a,\angle B=b, \angle C=c,\angle GCA=x$ Let $G$ be the intersection of $(ABC)$ and $(CEF)$, notice that $O$ belongs to $(CEF) \Leftrightarrow \angle OGE=\angle OCE= \angle OCB = 90-a$, and $\angle OGE=\angle BGE- \angle OGB=b-x - (90-x-c) = 90-a$ iff $BE=EG$. So notice we just need to prove $\triangle FEG \equiv \triangle FED$, and since we already have a common side $FE$ and a common angle ($\angle FGE=\angle FCE=c=\angle FDE=180-\angle FDA-\angle EDB=180-a-b$, which is true). Then, we'll prove $\angle FED=x$ to conclude by SAA congruence. To prove this we'll prove $\triangle FED\sim \triangle ABG$, but since $\angle EDF=\angle BGA$, I spent very much time on angle chase, but I couldn't do anything, then I thought: what if I tried the SAS similarity to prove it? Here it is: We need $\dfrac{DF}{DE}=\dfrac{AG}{BG} \Leftrightarrow \dfrac{AF}{BE}=\dfrac{AG}{BG} \Leftrightarrow \dfrac{BE}{BG}=\dfrac{AF}{AG}$, which is true because $G$ is the miquel point of $ABEF$, and there's a spiral similarity taking $BE \to AF$. $Q.E.D$

Let $(ABC)$ be the unit circle. $E\in BC\Rightarrow e+bc\bar{e}=b+c\Rightarrow\bar{e}=\frac{b+c-e}{bc}$ The midpoint of $BD$ lies on the perpendicular from $E$ to $AB$ $\Rightarrow a+b+e-ab\bar{e}=b+d\Rightarrow\bar{e}=\frac{a-d+e}{ab}$ Thus $\frac{b+c-e}{bc}=\frac{a-d+e}{ab}\Rightarrow e=\frac{ab+cd}{a+c}$ Analogously $f=\frac{ab+cd}{b+c}$. Also $D\in AB\Rightarrow \bar{d}=\frac{a+b-d}{ab}$ We have $\frac{e-o}{c-o}\cdot\frac{c-f}{e-f}=\frac{ab+cd}{c(a+c)}\cdot\frac{c^2+bc-ab-cd}{(b+c)(ab+cd)(\frac{1}{a+c}-\frac{1}{b+c})}=\frac{c^2+bc-ab-cd}{bc-ac}$ Thus $OECF$ is cyclyc $\iff$ $\iff \Bigl(\frac{c^2+bc-ab-cd}{bc-ac}\Bigr)=\overline{\Bigl(\frac{c^2+bc-ab-cd}{bc-ac}\Bigr)}\iff\Bigl(\frac{c^2+bc-ab-cd}{bc-ac}\Bigr)=\Bigl(\frac{\frac{1}{c^2}+\frac{1}{bc}-\frac{1}{ab}-\frac{a+b-d}{abc}}{\frac{1}{bc}-\frac{1}{ac}}\Bigr)\iff\frac{(a-b)}{abc}(c^2+bc-ab-cd)=c(b-a)\frac{(ab+ac-c^2-ac-bc+cd)}{abc^2}\iff c^2+bc-ab-cd=-ab-ac+ac+c^2+bc-cd\iff 0=0$ $Q.E.D.$