Actually replacing an $1 \times n$ rectangle by a $\sqrt{2} \times n$ rectangle will not change the answer (provided the sides of length $n$ are still divided into n equal segments). To make matters worse, we position the rectangle vertically, more or less like the Monolith of Space Odyssey fame. Naturally this position requires three coordinates to describe, so the vertices of the rectangle will be at $(1, 0, 0), (0, 1, 0), (1, 0, n),$ and $(0, 1, n).$ The marked points will have the coordinates $(1, 0, k)$ and $(0, 1, k)$ for $0 \ge k \ge n.$ For better view we establish ourselves at $(0, 0, -1)$ and project the marked points onto the plane $z = 0.$ The images of points $(1, 0, k)$ and $(0, 1, k)$ under this projection will be $(\frac{1}{k+1} , 0, 0)$ and $(0,\frac{1} {k+1} , 0).$ In other words, what we see in the plane $z = 0$ is the original triangle $ABC$ with its marked points except one small triangle in the corner. Thus our projection provides a bijection between all the parts of the rectangle and all but one parts of the triangle. This, of course, means that the number of parts in the triangle was greater by $1.$