Let us eliminate a corner case first:
If the $z_k$ lie on a line, we can presume that $z_2 = \lambda z_1+(1-\lambda) z_3$ with $\lambda \in (0,1)$. Then the trivial
triangle (and its boundary) is given by the segment $[z_1,z_3]$, and it is easy to see that
$\sum_k t_k z_k = (t_1 + \lambda t_2)z_1+(t_3+(1-\lambda)t_2) z_3$. Since
$t_1 + \lambda t_2\ge 0, t_3+(1-\lambda)t_2 \ge 0$ and $t_1 + \lambda t_2+t_3+(1-\lambda)t_2 =1$, we see that $\sum_k t_k z_k \in [z_1,z_3]$.
So, we can presume that the $z_k$ do not lie on a line.
The issue is how to define what we mean by a triangle and the inside of a triangle.
Some preliminaries first:
It is straightforward to
show that the points $\binom{1}{z_k}$ are linearly independent (over $\mathbb{R}$) and hence the equation $\sum_k \beta_k \binom{1}{z_k} = \binom{1}{z}$
has a unique solution for any $z \in \mathbb{C}$. The $\beta_k$ are called the
barycentric coordinates of $z$ (with respect to the $z_k$).
Define $\beta:\mathbb{C} \to \mathbb{R}^3$ as the unique solution to the above equation. That is, for any $z \in \mathbb{C}$ we have $\sum_k \beta_k(z) = 1$
and $z = \sum_k \beta_k(z) z_k$. It is not difficult to show that $\beta$
is continuous.
One definition is to define a triangle as the convex hull of the points $T = \operatorname{co} \{z_1,z_2,z_3\} \subset \mathbb{C}$, and define the inside as the topological interior.
Then we see that $z \in T$ iff $\beta_k(z) \ge 0$ for all $k$, and it follows that $T$ is closed since $\beta$ is continuous. Since $T$ is closed,
we have $T = T^\circ \cup \partial T$ and the desired result follows.
An alternative method is to define it as the appropriate intersection of
'half spaces' defined by, for example, a line through $z_1,z_2$ and a 'side', that is, the half space defined by the line that $z_3$ lies in (assuming
that $z_3$ is not on the line).
To define a half space, suppose $w_1,w_2,w_3$ are such that $w_1 \neq w_2$ and $w_3$ is not on the line through $w_1,w_2$.
Let $\sigma = \operatorname{sgn} (\operatorname{im} (\overline{(w_2-w_1)} (w_3-w_2))) $.
The half space can be described (rather clumsily) by
$H(w_1,w_2,w_3) = \{ w | \sigma \operatorname{im} (\overline{(w_2-w_1)} (w-w_1)) \ge 0 \}$. Continuity shows that the half space is closed.
It is not hard to see that the interior is given by
$H(w_1,w_2,w_3)^\circ = \{ w | \sigma \operatorname{im} (\overline{(w_2-w_1)} (w-w_1)) > 0 \}$.
Then define $T = H(z_1,z_2,z_3) \cap H(z_2,z_3,z_1) \cap H(z_3,z_1,z_2)$.
It is not hard to verify that
$T^\circ = H(z_1,z_2,z_3)^\circ \cap H(z_2,z_3,z_1)^\circ \cap H(z_3,z_1,z_2)^\circ$.
Choose $z=\sum_k \beta_k z_k$, where $\beta_k$ are the barycentric coordinates. Then we must show that $z$ lies in each of the half spaces.
Pick the first one. Let $\sigma = \operatorname{sgn} (\operatorname{im} (\overline{(z_2-z_1)} (z_3-z_2))) $.
Note that $z-z_1 = \beta_3 (z_3-z_1) + \beta_2 (z_2-z_1)$, and so
$\operatorname{sgn} (\operatorname{im} (\overline{(z_2-z_1)} (z-z_2))) = \beta_3 \sigma$. Hence we see that $z \in H(z_1,z_2,z_3)$ iff $\beta_3 \ge 0$
and $z \in H(z_1,z_2,z_3)^\circ$ iff $\beta_3 > 0$.
Now suppose $z = \sum_k t_k z_k$ as in the question, then we see that
$t_k = \beta_k$ by uniqueness and so $z \in T$, and as above we see that
this implies that $z$ either lies in the boundary or in the interior.
