Should it be $\angle E'C_1F' = \angle ACB$? Let $A',B'$ be points such that $A',B'$ are symetrical of $C$ wrt $A,B$. Now $CEA'E'$, $CFB'F'$ are parallelograms and notice that $C,C_1$ are symetrical wrt $A'B'$ so we get $\triangle E'C_1A', \triangle F'C_1B'$ are similar (easy angle chasing + $\triangle AEC, \triangle BFC$ are similar) $\implies \angle E'C_1F' = \angle A'C_1B'= \angle ACB$.

math4444 wrote: Should it be $\angle E'C_1F' = \angle ACB$? Thanks, edited.

We know $ABTC$ is cyclic hence by Ptolemy's theorem we have $AB\cdot sin(\angle FAC)+AC\cdot sin(\angle FAB)=AT\cdot sin(\angle BAC)$.We need toprove that $AY\cdot sin(\angle BAC)=AX\cdot sin(\angle FAC)+AZ\cdot sin(\angle FAB)$ thus it suffices to prove that $TY\cdot sin(\angle BAC)=CZ\cdot sin(\angle FAB)+BX\cdot sin(\angle FAC)$ suppose $\angle BAC=x,\angle FAB=y,\angle FAC=z$. $CZ=CE=sin (z)\cdot BC$ and simirally $BX=sin (y)\cdot BC$ and $TY=HT=2HF=2sin (y)\cdot HC=2sin (y)\cdot \frac{1}{sin (x)} \cdot CE=2sin (y)\cdot \frac{1}{sin (x)} \cdot sin (z)\cdot BC$ and we are done

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Mahdi.sh wrote: We know $ABTC$ is cyclic hence by Ptolemy's theorem we have $AB\cdot sin(\angle FAC)+AC\cdot sin(\angle FAB)=AT\cdot sin(\angle BAC)$.We need toprove that $AY\cdot sin(\angle BAC)=AX\cdot sin(\angle FAC)+AZ\cdot sin(\angle FAB)$ thus it suffices to prove that $TY\cdot sin(\angle BAC)=CZ\cdot sin(\angle FAB)+BX\cdot sin(\angle FAC)$ suppose $\angle BAC=x,\angle FAB=y,\angle FAC=z$. $CZ=CE=sin (z)\cdot BC$ and simirally $BX=sin (y)\cdot BC$ and $TY=HT=2HF=2sin (y)\cdot HC=2sin (y)\cdot \frac{1}{sin (x)} \cdot CE=2sin (y)\cdot \frac{1}{sin (x)} \cdot sin (z)\cdot BC$ and we are done Veeeeeerrrrrry nice solution.....thanks Mahdi.sh

(Hard to decide whether coordinate bash or complex bash is easier here.)

Assassino9931 wrote: (Hard to decide whether coordinate bash or complex bash is easier here.)

Here's a complex bash: Let $(ABC)$ be the unit circle. Then, by the feet of perpendicular formula

we have that: $$e=\frac{a^2+ab+ac-bc}{2a},\quad f=\frac{b^2+ba+bc-ac}{2b},\quad d=\frac{c^2+ca+cb-ba}{2c}$$$$e'=2a-e=\frac{-ab+bc+3a^2-ac}{2a}$$$$f'=2b-f=\frac{-ab+ac+3b^2-bc}{2b}$$$$c_{1}=4d-3c=\frac{2ca+2cb-2ab-c^2}{c}$$Now $\angle E'C_{1}F'=\angle ACB\iff \frac{e'-c_{1}}{f'-c_{1}}\Big/\frac{b-c}{a-c}\in\mathbb{R}$, so we bash:

$$\frac{e'-c_{1}}{f'-c_{1}}\Big/\frac{b-c}{a-c}=\frac{\frac{-ab+bc+3a^2-ac}{2a}-\frac{2ca+2cb-2ab-c^2}{c}}{\frac{-ab+ac+3b^2-bc}{2b}-\frac{2ca+2cb-2ab-c^2}{c}}\Big/\frac{b-c}{a-c}=$$$$=\frac{b(a-c)(4ab-ac-bc))}{a(b-c)(4ab-ac-bc)}\Big/\frac{b-c}{a-c}=\frac{b(a-c)^2}{a(b-c)^2}$$And obviously

$\frac{b(a-c)^2}{a(b-c)^2}\in\mathbb{R}$, so we're done!

We complex bash. Note that \[d = \frac12 (a+b+c- \frac{ab}{c}), e = \frac12 (a+b+c-\frac{bc}{a}), f = \frac12 (a+b+c-\frac{ac}{b})\]Then, \[e' = 2(a-e) + e = \frac12 (3a-b-c+\frac{bc}{a}) \]\[f' = 2(b-f)+f = \frac12 (-a+3b-c+\frac{ac}{b})\]\[c_1 = 4(d-c) + c = \frac12 (4a+4b-2c-\frac{4ab}{c})\]Now, we note that \[\frac{\frac{e'-c_1}{f'-c_1}}{\frac{b-c}{a-c}} = \frac{-a-5b+c+\frac{bc}{a} + \frac{4ab}{c}}{-5a-b+c + \frac{ac}{b} + \frac{4ab}{c}}\cdot \frac{a-c}{b-c}\]And for the first time ever, directly taking the conjugate screws stuff up too much. So, we factor first. Motivated by hoping that the top is divisible by $b-c$, noticing that it fails, but noticing that $a=c$ cancels the $-5b$ and $\frac{4ab}{c}$, we have \[= \frac{\frac{1}{ac} (a-c)(4ab-bc-ac)}{\frac{1}{bc} (b-c) (4ab-ac-bc)}\cdot \frac{a-c}{b-c} = \frac{b(a-c)^2}{a(b-c)^2}\]Taking the conjugate, we get \[\frac{\frac1b (\frac1a - \frac1c)^2}{\frac1a (\frac1b- \frac1c)^2} = \frac{a^2b^2c^2}{a^2b^2c^2} \frac{b (c-a)^2}{a(c-b)^2}\]Thus, since taking the conjugate kept the expression equal, that means that $\frac{\frac{e'-c_1}{f'-c_1}}{\frac{b-c}{a-c}}$ is real, and $\angle E'C_1F' = \angle BCA$ and we're done. $\blacksquare$.