I am talking about triangles in both 2D and 3D. I feel the statement is true, but could not come up with a proof.
The bounding box is the simplest axis-aligned type. Below are some 2D examples, regardless of the triangle shape, the centroid of its bounding rectangle (intersection of the two diagonals) falls within the triangle -- note that I consider point-on-the-edge as point-within. Are there exceptions where the rectangle's centroid falls outside of the triangle it bounds?
In 3D things get more complicated, one needs to show that the centroid of the bounding box lays on the plane of the rectangle, which may or may not be tree.
Any suggestion is welcomed!
The equations of the lines that make the boundary of the bounding box are $$x=\min\{x_i,i=1,2,3\}$$$$x=\max\{x_i,i=1,2,3\}$$$$y=\min\{y_i,i=1,2,3\}$$$$y=\max\{y_i,i=1,2,3\}$$
Three vertices, four values, therefore, by the pigeonhole principle:
This is clear from the picture, but now we have a proof.
Without loss of generality assume that the corner is $A_3=(x_3,y_3)=(\max\{x_i\},\max\{y_i\})$. The point corresponding to $\min \{x_i\}$ lies somewhere on the left wall (call it $A_1$), and the point corresponding to $\min \{y_i\}$ lies somewhere on the bottom side (call it $A_2$). The triangle is the intersection of the half planes:
"South East of $A_1A_3$" $\cap$ "North West of $A_2A_3$" $\cap$ "North East of $A_1A_2$".
If you take individually the sharpest situation for all three half planes, the intersection still contains the centre of the rectangle.