Suppose the polygon has vertices , , ... , , listed in clockwise order. Then the area () of is
You can also go counterclockwise order, as long as you find the absolute value of the answer.
The Shoelace Theorem gets its name because if one lists the coordinates in a column, and marks the pairs of coordinates to be multiplied,
for(int i=1; i < 3; ++i; label("$i$",(0,-i)); /* draw(graph(F(i*pi),0,1),Pen(i), "$\sin("+(i == 1 ? "" : (string) i)+"\pi x)$"); */ (Error compiling LaTeX. for(int i=1; i < 3; ++i; ^ afc1be4824a9f80725a9496c36f09f1d1c545d39.asy: 5.24: syntax error error: could not load module 'afc1be4824a9f80725a9496c36f09f1d1c545d39.asy')
the resulting image looks like laced-up shoes.
This can also be written in form of a summation And thus we can introduce determinants to get This is more helpful in the formula variant of the Shoelace theorem.
Claim 1: The area of a triangle with coordinates , , and is .
Proof of claim 1:
Writing the coordinates in 3D and translating so that we get the new coordinates , , and . Now if we let and then by definition of the cross product .
We will proceed with induction.
By claim 1, the shoelace theorem holds for any triangle. We will show that if it is true for some polygon then it is also true for .
We cut into two polygons, and . Let the coordinates of point be . Then, applying the shoelace theorem on and we get
Let be the set of points belonging to the polygon. We have that where . The volume form is an exact form since , where Using this substitution, we have Next, we use the theorem of Stokes to obtain We can write , where is the line segment from to . With this notation, we may write If we substitute for , we obtain If we parameterize, we get Performing the integration, we get More algebra yields the result
This is a very nice approach that directly helps in understanding the sum as terms which are areas of trapezoids.
See page 281 in this book (in the Polygon Area section.) https://cses.fi/book/book.pdf
(The only thing that needs to be modified in this proof is that one must shift the entire polygon up by k, until all the y coordinates are positive, but this term gets canceled in the resulting sum.)
In right triangle , we have , , and . Medians and are drawn to sides and , respectively. and intersect at point . Find the area of .
A good explanation and exploration into why the theorem works by James Tanton:  AOPS