It's clear that it's enough to prove $\frac{TC}{TB}=\frac{EC^2}{BD^2}$ and denoting $U,V$ the intersection of the tangent circle and $BC$ w.t.p. that $U$ lies between $B$ and $V$ and using the fact that $BD^2=BU \cdot BV, CE^2=CV \cdot CU$ and that $TU \cdot TV= TB \cdot TC= TK \cdot TL$, our statement becomes equivalent to $\frac{TC}{TB}=\frac{CV \cdot CU}{BU \cdot BV}$, which is easy to prove taking into account that $TB \cdot TC= TU \cdot TV$ (i just denoted $TB=x, UB=y, UV=z, VC=t$, both relations being equivalent to $xt=y(x+y+z))$.

MariusBocanu wrote: It's clear that it's enough to prove $\frac{TC}{TB}=\frac{EC^2}{BD^2}$ $. can you explain here in detail please,

epsilon07 wrote: MariusBocanu wrote: It's clear that it's enough to prove $\frac{TC}{TB}=\frac{EC^2}{BD^2}$ $. can you explain here in detail please, need the explanation anymore??

to prove $\frac{TC}{TB}=\frac{EC^2}{BD^2}$ use \[ \frac{AX}{AB}=\frac{CE}{BD+CE} \quad \text{and} \quad \frac{AY}{AC}=\frac{BD}{BD+CE} \] and menelaus theorem to triangle ABC and point B,X,A

//T denotes the intersection of XY,BC the main point is to prove that TB* TC=TU*TV , then T lies on radical axis of cir(ABC) and the tangent circle i.e. the radical axis LK pass through T.

but i'm still confuse why $\frac{TC}{TB}=\frac{CV \cdot CU}{BU \cdot BV}$ is equivalent to $xt=y(x+y+z)$