According to the assumptions of the problem, we have $\angle ADT = \angle TKC$ and $\angle TCK = \angle TDK = \angle TAD$. From these two equations, it follows that $\triangle TDA \sim \triangle TKC$. Therefore, we can write:
$$\frac{KC}{AD} = \frac{TC}{AT} \qquad (1)$$
On the other hand, if we denote the midpoint of $BC$ as $M$, we can easily deduce that $\triangle ABM \sim \triangle ATC$. Therefore, according to 1, we have:
$$\frac{KC}{AD} = \frac{TC}{AT} = \frac{AB}{BM} = \frac{AB}{CM} \qquad (2)$$
We have shown that $\angle TCK = \angle TAD = \angle TAC$, which is equivalent to the fact that $KC$ is tangent to the circumcircle of triangle $ABC$. Thus, $\angle BAD = \angle MCK$. Using this equality and 2, we obtain $\triangle MCK \sim \triangle BAD$, which means $MK = MC$. Thus, in triangle $BCK$, the $K$-median is half of $BC$, and we must have $\angle BKC = 90^\circ$.