Quite sure ? It seems you made a mistake, you didn’t define $N$ .

Thanks omarius. Fixed.

jestrada wrote: Let $\Delta ABC$ be a triangle inscribed in the circumference $\omega$ of center $O$. Let $T$ be the symmetric of $C$ respect to $O$ and $T'$ be the reflection of $T$ respect to line $AB$. Line $BT'$ intersects $\omega$ again at $R$. The perpendicular to $CT$ through $O$ intersects line $AC$ at $L$. Let $N$ be the intersection of lines $TR$ and $AC$. Prove that $\overline{CN}=2\overline{AL}$. We have that $\angle NTC=\angle RTC=\angle T'BC=\angle B-\angle ABT'=\angle B-\angle ABT=\angle B-\angle ACO=2\angle B-90^\circ$. Also, $\angle AOC=2\angle B \Rightarrow 2\angle B=\angle AOL+\angle LOC=\angle AOL +90^\circ \Rightarrow \angle AOL=2\angle B-90^\circ=\angle NTC \Rightarrow \angle AOL=\angle NTC$ (1). Also, $\angle NCT=\angle ACO=\angle OAL \Rightarrow \angle NCT=\angle OAL$ (2). From, (1), (2) we have that triangles $NCT$ and $AOL$ are similar, so $\dfrac{AO}{TC}=\dfrac{AL}{NC} \Rightarrow \dfrac{AL}{NC}=\dfrac{AO}{2AO} \Rightarrow TC=2AO$.

Solution. Observe that $\bigtriangleup CLT$ is isosceles at $L$. Since $\angle RBA=\angle T'BA=\angle TBA$, we also have that $RA=AT$, hence $$\angle LTC=\angle LCT=\angle ACT=\angle TRA=\angle ATR=\angle ATN$$i.e. $LT$ and $NT$ are isogonals of $\bigtriangleup ATC$; thus, according to Sterner's theorem, we have $$\frac{AN}{NC}\cdot\frac{AL}{LC}=\frac{AT^2}{CT^2}\Longrightarrow \frac{AL}{NC}=\frac{AT^2}{CT^2}\cdot\frac{LT}{AN}$$Let $\angle ACT=\alpha$. The right-hand side of the last equality corresponds to $$\frac{\sin^2\alpha}{\sin 2\alpha \tan \alpha}=\frac{\sin^2 \alpha}{2\sin\alpha \cos \alpha }\frac{\cos \alpha}{\sin \alpha}=\frac{1}{2}$$therefore, $CN=2AL$, as required. $\blacksquare$

Let $\angle TBA=\angle T'BA=\angle TCA=\angle RCA=\frac{1}{2}\angle ALT=x$, then easy to see that, $A$ is the midpoint of arc $TR$ not containing $C$, hence, $\Delta ALT \sim \Delta RCT$ and $\Delta RCN \sim \Delta ATN$, using which we have, $$\frac{AL}{RC}=\frac{LT}{CT}, \text{ and } \frac{RC}{AT}=\frac{CN}{TN} \Longrightarrow \frac{AL \cdot CT}{LT}=\frac{CN \cdot AT}{TN} \Longrightarrow \frac{AL \cdot 2OC}{\tfrac{OC}{\cos x}}=CN \cos x \implies \boxed{2AL=CN}$$

Easy complex bash Let $\omega$ be unit circle so that $|a|=|b|=|c|=1$ and $t=-c$ Then $t'=\frac{(a-b) \overline{t} + \overline{a}b - a\overline {b}}{\overline{a}-\overline{b}}=\frac{\frac{b-a}{c}+\frac{b}{a}-\frac{a}{b}}{\frac{1}{a}-\frac{1}{b}}=\frac{ab+bc+ca}{c}$ and easily note that $\overline{t'}=\frac{a+b+c}{ab}$ Since $T'$ lies on $BR$ we have $t'+br\overline{t'}=b+r \implies r=\frac{b-t'}{b\overline{t'} -1}\implies r=-\frac{a^2}{c}$(easy computing) Let us now compute $n$ $n=\frac{tr(a+c)-ac(t+r)}{tr-ac}=\frac{a^2(a+c)+ac(c+\frac{a^2}{c})}{a^2-ac}=\frac{2a^2+ac+c^2}{a-c}$ and note that $\overline{n}=\frac{2c^2+ac+a^2}{ac(c-a)}$ Since $L$ lies on $AC$ we have $l+ac\overline{l}=a+c\implies \overline{l}=\frac{a+c-l}{ac}$ On the other hand since $O$ is projection of $L$ onto $TC$ we have $\frac{1}{2}(l+t+c-tc\overline{l})=0 \implies \overline{l}=\frac{l}{tc}$ Thus $\frac{a+c-l}{ac}=\frac{l}{tc} \implies l=\frac{c(c+a)}{c-a}$ and note that $\overline{l}=\frac{c+a}{c(a-c)}$ Now we need to show $NC=2LA \implies (n-c)(\overline{n}-\overline{c})=4(l-a)(\overline{l}-\overline{a})$ After more computation we get $LHS=-\frac{(2a^2+2c^2)^2}{ac(c-a)^2}$ and $RHS=-4\times \frac{(a^2+c^2)^2}{ac(c-a)^2}$ so we are done