I have appropriated the shoelace and R. Nelsen's proofs but not the last one, which seems to me directly related to a number of proofs:

#2, #10, #24

Roger notes that this diagram is also closely (if not exactly) associated with proof #27.

Indeed many of these proofs are variants of each other. I only included the diagram because this is how I saw the areas being played out from the shoelace rather than Roger's image (which although equivalent makes much more sense in the context of that formula).

Still I wonder how one derived Gauss area formula.

https://www.avmathteam.org/lessons/shoelacetheorem

And then asked Roger and he sent me this:

https://www.artofproblemsolving.com/blog/25360

(likely copied onto the one I found)

The two seem to be identical and avoid the use of the PT in order to derive the Shoelace Formula.

Still I happen to like the fact that the formula is a simpler proof than many of the other ones I have done.

Perhaps someday I will contribute something outside of the PT...

A very nice proof!

--Stuart Anderson

On the page for proof #91, you wrote:

"A question could be asked whether the Shoelace formula, in itself, is based on the Pythagorean theorem, causing a vicious circle in proving the latter."

Is this still in question, given Stuart's comment in message #11?

https://www.cut-the-knot.org/pythagoras/proof91.shtml

You have written "...the are of the square formed..." where I think you mean to say "area"

c^2 = (a/c)cosB - cos(B-C)/bc

Quite surprising this could be as c^3 = ....

c^2 = (a/c)cos(B) - cos(B+C)/bc

I forgot about the sign change in the sum formula for cosine.

I think there is still light at the end of the tunnel if one rewrit's B = pi - (A+C) and B+C = pi - A and the fact that cos(x) = -cos(pi-x) for 0 < x < 180.

I'll keep you 'posted.'

The whole thing is futile though because it produces the identity:

c = a*cosB + b*cosA

Which could be gained much easier by drawing an altitude...

So if you ever needed to prove something using this identity without using the sine rule (which is drawing an altitide), I suppose you get at it using the shoelace formula.

If it matters to anyone, here are THE CORRECT vertices of the square:

(0,a)
(b*sinC, b*cosC)
(b*sinC + c*cosB, b*cosC + c*sinB)
(c*cosB, a + c*sinB)

I think I have to be a bit more precise about this last post.

First of all, some background considerations about determinants:
The 3) basic properies of determinants:
a) det(coordinate basis) = 1
b) det( ..., k·u, ...) = k·det(..., u , ...)
c) det(..., u , ... , v , ...) = det(... , u , ... , v + k u , ...) = det(... , u + k v , ... , v , ...)
easily yield the following three more properties:
a') det( ... , u , ... , u , ...) = 0
b') det( ... , u , ... , v , ...) = - det( ... , v , ... , u , ...)
c') det( ... , u + u' , ...) = det( ... , u , ...) + det( ... , u' , ...)

Now, consider the following geometrically intuitive hypotheses:
i) for any positive integer n, each coordinate vector in R^n made of only one component equal to 1 and all other components equal to 0 is a unitary vector in R^n
ii) for any positive integer n, each coordinate vector in R^n along the axis x_i is orthogonal to any of the vectors in the subspace x_i=0
iii) for any positive integer n, a unitary vector in R^n yields a unitary vector of R^(n+1) by adding a 0 component in some position as (n+1)th component. By this adding of 0, also orthogonal vectors of R^n yield orthogonal vectors of R^(n+1).
iv) if u and v are unitary mutually orthogonal vectors, then the vectors h·u+k·v and (-k)·v+h·u (both vectors of the (u,v)-subspace) are mutually orthogonal, and if h·u+k·v then also (-k)·v+h·u is unitary
v) if a vector is orthogonal to some vectors, it is also orthogonal to a linear combination of these vectors
vi) for any positive integer n, any n-tuple of unitary vectors in R^n gives a determinant having 1 as absolute value.

If i)...vi) are true, then, for any positive integer n, every unitary (i.e of length 1) vector u=(u_1,...,u_n) in R^n has the two following properties:
1) u can be "completed" with n-1 more unitary mutually orthogonal vectors v_1,...,v_(n-1), all orthogonal to the vector u, to build a n-dimensional determinant det( u , v_1 , ... , v_(n-1) ) = 1
2) Σ(u_i)² = 1 (i through 1,...,n)

Proof:
By induction suppose that in R^n every unitary (i.e of length 1) vector u=(u_1,...,u_n) has the two following 1) and 2).
( this is true for n=2, where v = (- u_2 , u_1) )

Consider a unitary vector a=(a_1,...,a_(n+1)) in R^(n+1) and suppose, that it is not along the fist axis x_1 (i.e. a_2,...,a_(n+1) are not all equal to 0; otherwise a would be the first coordinate vector and the properties 1) and 2) would trivially flollow) and that s=a_1 is non null (otherwise we could treat a as an n-dimensional vector).
So, the "unheaded" vector a*=(a_2,...,a_(n+1)) is a non-zero vector of R^n.
Put r=√<Σ(a_i)²> (i through 2,...,n+1) and u=a*/r, so that u is a unitary vector of R^n and the inductive hypotheses are verified.
Let's call v_1,...,v_(n-1) the n-1 completing unitary vectors of R^n given by property 1) and let's call v*_1,...,v_*(n-1) the same vectors with a zero prefixed as first coordinate (so they are all vectors of R^(n+1).
The (n+1)-dimensional vectors (0,a_1,...,a_n) (let's denote it by the same symbol a*) the vector u*=(0,u_1,...,u_n) and the completing vectors v*_1,...,v_*(n-1) are all in the coordinate (n+1)-dimensional subspace x_1=0 and the vector a is a linear combination of u* and the (n+1)-dimensional vector w having all null components but the first one which is equal to 1; actually, it is: a=a_1·w+a**=s·w+r·u*, u* and w being unitary mutually orthogonal vectors,
so that the vector a'=-r·w+s·u* is orthogonal to the vector a.
It is: det( w , u* , v*_1 , ... , v_*(n-1) ) = det( u , v_1 , ... , v_(n-1) ) = 1.
Now we consider the vectors a' and v*_1 , ... , v_*(n-1) as the n "completing" mutually orthogonal unitary vectors of the unitary vector a (all of them are orthogonal to the vector a).
We also have:
1 = det( a , a' ,v*_1 ,... , v_*(n-1) ) = det( s·w+r·u*, -r·w+s·u* , v*_1 ,... , v_*(n-1) ) =
= det( s·w , -r·w+s·u* , v*_1 ,... , v_*(n-1) ) + det(r·u*, -r·w+s·u* , v*_1 ,... , v_*(n-1) ) =
= s²·det( w , u*, v*_1 ,... , v_*(n-1) ) + r²·det(u* , - w , v*_1 ,... , v_*(n-1)) = s²+r²
so that 2) is verified.

The best way to understand this process is to develop the step from n=2 to n+1=3. The induction could also be started, in a more subtle and elegant way, at n=1 instead of n=2.