With this kind of problems it is very important to draw a big figure and try to see if there is anything we can say about the common intersection. First, we notice that the centres of $K_A$, $K_B$ and $K_C$ lie on the incircle $K$ of the triangle $ABC$.

Let $A_2$ be the point where the bisector of the angle $B_1A_1C_1$ intersects $K$ for the second time ($A_2$ is the midpoint of the arc $B_1C_1$). Then $\overline{B_1C_1A_2} = \angle B_1A_1A_2 = \angle A_2B_1C_1$, and the tangent-chord theorem implies that $\angle A_2C_1A = \angle A_2B_1C_1$. So, $\angle B_1C_1A_2 = \angle A_2C_1A$ and $C_1A_2$ is the bisector of the angle $B_1C_1A$. On the other hand, $AA_2$ is the bisector of the angle $\angle BAC$, so $A_2$ is the incentre of the triangle $AC_1B_1$.

As we draw the line $A_1A_2$ onto the figure we notice that the common intersection of the tangents lies on this line. Let $B_2$ and $C_2$ be the centres of $K_B$ and $K_C$. If we also draw $B_1B_2$ and $C_1C_2$, we see that they contain the common intersection of the tangents as well. The lines $A_1A_2$, $B_1B_2$ and $C_1C_2$ are the bisectors of the inner angles of the triangle $A_1B_1C_1$ and they meet in a point we denote by $J$. We wish to prove that each of the tangents $t_A$, $t_B$ and $t_C$ also contains $J$.

Since $J$ is the intersection of the angle bisectors of the triangle $A_1B_1C_1$, we have
$$\angle C_2B_2J = \angle C_2B_2B_1 = \angle C_2C_1B_1 = \angle A_1C_1C_2 = \angle A_1B_2C_2$$
and
$$\angle JC_2B_2 = \angle C_1C_2B_2 = \angle C_1B_1B_2 = \angle B_2B_1A_1 = B_2C_2A_1$$

Hence, the triangles $A_1C_2B_2$ and $JC_2B_2$ are similar. They also have a common side, so they are congruent. Reflect the outer tangent $BC$ to the circles $K_B$ and $K_C$ in the line $B_2C_2$ connecting the two centres. We have just shown that this reflection maps $A_1$ to $J$. It also maps the tangent $BC$ into a tangent passing through the image of $A_1$. This implies that $J$ lies on $t_A$. Similarly, we show that $J$ lies on $t_B$ and $t_C$. We conclude that the three tangents intersect at a single point.

Let us find another way of describing the tangent $t_A$. We have noticed that it contains $J$. The figure also suggests that $t_A$ is parallel to $B_1C_1$. Let $l$ be the line through $J$ parallel to $B_1C_1$. We wish to show that $l$ is tangent to $K_B$ and $K_C$.

Let us draw a less cluttered figure. We will not need the circles $K_A$ and $K_C$. Let $D$ be the point on $l$ such that $B_2D$ is perpendicular to $l$. We wish to show that $|B_2D|$ is the diameter of $K_B$.

First, let us find the angle $\angle B_2JD$. Since $JD$ is parallel to $B_1C_1$, we have $\angle B_2JD = \angle B_2B_1C_1 = \angle B_2A_1C_1$. Let $E$ be the midpoint of $A_1C_1$. We wish to show that $|B_2E| = |B_2D|$. We have $\angle B_2JD = \angle B_2A_1E$, so the triangles $JDB_2$ and $A_1EB_2$ are similar. We wish to show that they are congruent, so $|A_1E| = |JD|$ or $|A_1B_2| = |JB_2|$. The second equality is equivalent to the fact that $B_2A_1J$ is an isosceles triangle with the apex at $B_2$. Let $\angle A_1C_1B_1 = \gamma$. Then
$$\angle B_2JA_1 = \pi - \angle A_1JB_1 \Rightarrow \pi - \left(\frac{\pi + \gamma}{2}\right) = \frac{\pi - \gamma}{2}.$$
Clearly, $\angle JB_2A_1 = \angle B_1B_2A_1 = \angle B_1C_1A_1 = \gamma$, so the third angle in the triangle is equal to $\angle A_1B_2 = \frac{\pi - \gamma}{2}$. The triangle $B_2A_1J$ is isosceles with the apex at $B_2$, so $|A_1B_2| = |JB_2|$ and the triangles $JDB_2$ and $A_1EB_2$ are congruent. So, $|B_2D| = |B_2E|$ and $JD$ (or $l$) is tangent to $K_B$. Similarly, we show that $l$ is tangent to $K_C$, so $l = t_A$. The same arguments would show that $t_B$ is the line through $J$ parallel to $A_1C_1$ and $t_C$ is the line through $J$ parallel to $A_1B_1$. Hence, all three tangents intersect at a single point, namely $J$.