Taiwan IMO Selection Camp

Without loss of generality, assume $AB > AC$. Construct the circumcircle $\omega$ of $\triangle AEF$.

Lemma 1. The line $ME$, the line $MF$, and the line $AT$ are all tangent to circle $\omega$.

Proof. Note that $E$, $F$ both lie on the circle with diameter $BC$, whose center is $M$. Hence $MC = ME = MF = MB$. Therefore
$$\begin{aligned} \angle FEM &= \angle FEB + \angle BEM = \angle FCB + \angle MBE \\ &= (90^\circ - \angle CBF) + (90^\circ - \angle ECB) \\ &= 180^\circ - \angle CBA - \angle ACB = \angle FAE. \end{aligned}$$
So by the tangent-chord angle property, $ME$ is tangent to $\omega$. By symmetry, $MF$ is also tangent to $\omega$.

Moreover, $\angle TAB = \angle CBA = 180^\circ - \angle BEC = \angle AEF$, so $TA$ is also tangent to $\omega$. ♡

Now consider the circle $\omega$, together with the circle $\gamma$ centered at $M$ with radius $0$. Since $KE^2 = KM^2$, and $KE$ is tangent to $\omega$ while $KM$ is tangent to $\gamma$, $K$ lies on the radical axis of $\omega$ and $\gamma$. Similarly, so does $L$. In other words, the line $KL$ is exactly the radical axis of $\omega$ and $\gamma$, and $T$ lies on this radical axis. Hence, considering the power of $T$ with respect to these two circles: $TA^2 = TM^2$, which gives $TA = TM$. This completes the proof.