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Using only precalculus mathematics (including that the area of the triangle with vertices at the origin, $(x_1,y_1)$, and $(x_2,y_2)$ is half of the absolute value of the determinant of the $2\times 2$ matrix of the vertices $(x_1,y_1)$ and $(x_2,y_2)$, $\frac{1}{2}\cdot\left|x_1\cdot y_2 - x_2\cdot y_1\right|$) how can one prove that the shoelace method works for all non-self-intersecting polygons?

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up vote 8 down vote accepted

One way is to note that $x_1y_2 - x_2y_1$ is a signed area, i.e. it may positive or negative. Adding up all the signed areas of the triangles formed by the points $O$, $P_k$ and $P_{k+1}$ will cancel all the superfluous parts, as can be seen from the sketch:

Adding signed areas for the sides of a polygon

The same argument also works for trapezoids with the $x$-axis instead of triangles with the origin. (I think this is the most illuminating argument, because the key trick is to give the area a sign depending on orientation.)

One could argue that this is not very rigorous, however. A more rigorous proof is to divide the polygon into two smaller polygons (it's not trivial to show that this is possible) and argue that adding the shoelace sums of the two parts gives the shoelace sum of the whole. That's because the two additional terms for the extra side cancel each other. (This cancellation is well-known from line integrals, we are in essence calculating $\frac12 \oint_{polygon} x\,dy - y\,dx$ here.) By induction, you then only have to verify the formula for a triangle.

Shoelace sums of two polygons

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