In-center of a triangle

This may be similar enough to what you're looking for.

Represent the points as vectors $A=(x_a,y_a)$, $B=(x_b,y_b)$, $C=(x_c,y_c)$. Let a point $(p:q:r)$ stand for $$\frac{pA+qB+rC}{p+q+r},$$ so that $A=(1:0:0)$, $B=(0:1:0)$, $C=(0:0:1)$; we want to show that the incenter is $(a:b:c)$. (These are called barycentric coordinates, and the ``extra dimension,'' using three coordinates instead of just two, comes from the fact that if you scale $p$, $q$, and $r$ each by the same amount, the point it represents doesn't change.)

It shouldn't be too hard to convince yourself that lines can, in this system, be represented by a triple $(\alpha,\beta,\gamma)$, and consist of the points $(p:q:r)$ with $\alpha p+\beta q+\gamma r=0$. You can see this by considering the places in which a line intersects each of the sides of the triangle.

Now, let's find the equation of the bisector of $\angle BAC$. We know this line contains $A$, and it also contains the point $D$ on $BC$ with $\frac{BD}{DC}=\frac cb$. This point can be represented as $(0:b:c)$. So, the bisector of $\angle BAC$ is the line $0p+cq-br=0$, or equivalently $\frac qb=\frac rc$.

By symmetry, the three angle bisectors intersect at the point $(p:q:r)$ with $\frac pa=\frac qb=\frac rc$, i.e. the point $(a:b:c)$.

A side-note: You mention looking for a way to show this using determinants. While this isn't as direct, one can show that area in barycentric coordinates is closely related to determinants, in that if you have three points $(p_1:q_1:r_1)$, $(p_2:q_2:r_2)$, and $(p_3:q_3:r_3)$, normalized such that $$p+1+q_1+r_1=p_2+q_2+r_2=p_3+q_3+r_3=1,$$ then the area of the triangle represented by these three points is $$\begin{vmatrix}p_1&q_1&r_1\\p_2&q_2&r_2\\p_3&q_3&r_3\end{vmatrix}$$ times the area of $ABC$. One way to define the incenter is as the unique point $I$ inside $ABC$ for which $$\frac{\operatorname{Area}(IBC)}a=\frac{\operatorname{Area}(ICA)}b=\frac{\operatorname{Area}(IAB)}c.$$ Applying this barycentric area formula gives you the same result as before, i.e. that $I=(a:b:c)$. I believe this proof can, with enough work, be translated out of barycentric coordinates and into Cartesian, using the determinant version of the shoelace formula.