Complex numbers for $a$, $b$, $c$ lying on the unit circle with circumcenter $O=0$. $m=\frac{b+c}{2}$, $e=\frac{1}{2}(a+b+c-\frac{ca}{b})=\frac{ab+b^2+bc-ca}{2b}$, $f=\frac{1}{2}(a+b+c-\frac{ab}{c})=\frac{ca+bc+c^2-ab}{2c}$. $k=\frac{1}{2}(m+e)=\frac{2b^2+2bc+ab-ca}{4b}$, $l=\frac{1}{2}(m+f)=\frac{2c^2+2bc+ca-ab}{4c}$. $AT\parallel BC \implies \frac{t-a}{b-c}=\big( \overline{\frac{t-a}{b-c}}\big) \implies t-a=bc(\frac{1}{a}-\overline{t}) \implies \overline{t}=\frac{a^2+bc-at}{abc}$. $T \in KL \implies \frac{t-k}{t-l}=\big( \overline{\frac{t-k}{t-l}}\big) \implies t=\frac{4a^3+ac^2+ab^2+2abc-2a^2b-2a^2c-2bc^2-2b^2c}{4(a^2-bc)}$. $t-a=\frac{(b+c-2a)(ab+ca-2bc)}{4(a^2-bc)}$, $t-m=\frac{a(b+c-2a)^2}{4(a^2-bc)}$. Now it's very easy to see that $|t-m|^2=|t-a|^2$, as desired. $\square$