Let $P$ be the tangency point of the inscribed circle of $\triangle ABC$ with the side $BC$. Let $x = BK = BP$, $y = AK = AL$, $z = CP = CL$.
If $BC = AC$, then the statement of the problem holds because the points $M$, $K$, $T$ are coincide and $CT$ is a bisector of the angle $ACB$.

Fig. 1

Fig. 2

Let $AC > BC$ (see Fig. 1). Since $MN$ is the midline we have $MN \parallel AC$ and so $\angle KTM = \angle KLA$. But $\angle AKL = \angle KLA$ (whence $AK = AL$),
therefore $\angle MKT = \angle AKL = \angle KTM$, and
$$MT = MK = 0.5 \cdot AB - BK = 0.5(x + y) - x = 0.5(y - x).$$
Thus, $TN = MN - MT = 0.5(y + z) - 0.5(y - x) = 0.5(z + x) = CN$. Therefore $\triangle TNC$ is an isosceles triangle $\angle NTC = \angle NCT$. Since $MN \parallel AC$ we have $\angle NTC = \angle TCA$. From these two equalities we obtain $\angle NCT = \angle TCA$, i.e. $CT$ is the bisector of $\angle BCA$.

Similar consideration can be applied for the case $AC < BC$ (see Fig. 2).