Is the following conjecture correct?



> 
> **Conjecture:** 
> 
> 
> If $A,B,C,D$ are four points in general position in the euclidean plane, with
> 
> 
> $a:=\|C-B\|,\ \ b:=\|C-A\|,\ \ c:=\|B-A\|$  
> 
> $a':=\|D-A\|,\ b':=\|D-B\|,\ c':=\|D-C\|$ , 
> 
> 
> $\begin{align}
> a+b+c\ &>\ a+b'+c',\ a'+b+c',\ a'+b'+c\\ 
> c+c'\ &>\ a+a',\ b+b'\\
> \end{align}
> $, 
> 
> 
> then $A,B,C,D$ are in strict convex configuration iff 
>  $$(a'+b')((a+c')(a-c')-a'b')+b'(b+a)(b-a) \ \lt 0$$
> 
> 
> 


If so, then checking planar convexity of four points is possible with 6 length measurements and 5 multiplications. This checks convexity via an inequality, so it is stable under small displacements of the points into 3-space.


By contrast, using the equality


$$area(ABC) + area(ACD) = area(ABD) + area(BCD)$$


in a check of convexity, which also can be done with only six length measurements, makes the check unstable under those small displacements. The first approach is based on [Stewart's Theorem](https://en.wikipedia.org/wiki/Stewart%27s_theorem), where the lengths $a',b',c'$ are known as $d,m,n$; the second approach is based on [Heron's Formula](https://en.wikipedia.org/wiki/Heron%27s_formula) instead. 


  



**Addendum:** 


A week has passed, since I asked for the correctness of checking planar convexity with Stewart's formula without any reply in that respect.


I will therefore provide the requested "justification" and motivation for checking planar convexity by means of distances albeit that task seems simple in coordinate geometry. 


The first remark in that respect is a word of warning when checking convexity via sums of triangle areas. 


If the triangles aren't carefully choosen, one might get false results even with planar coordinates and $3\times 3$ determinants in the standard way as described by David Eppstein:  

Suppose you are given 4 points $A, B, C, D$ in general position in the Euclidean plane and, from those you calculate the triangle areas $ABC,ABD,BCD,ACD$; how do you then pair up the triangles to get a correct convexity check?
As it turns out, it may be necessary to make three checks for equality of sum of pairs of triangle areas. In contrast, it suffices to check whether the sum of the three smallest areas exceeds the largest area, in which case the points are definitely in strict convex configuration. 


Now consider the case where $A, B, C, D$ are in general configuration in a hyperplane of $E^n$; in that case the computational effort increases with the dimension of space. Another problem with point-coordinates is that they need not necessarily be cartesian, e.g. GPS coordinates, whence you have to convert them to 3D cartesian coordinates only to be able to calculate the triangle areas you need to know.  

If in contrast the 6 distances between the pairs of points have been measured, deciding about the planar convexity of the point's configuration via Stewart's formula would seem much easier (provided its correctness, which isn't confirmed yet).


And, not to forget that coordinates are no physical entities whereas distances are; so, albeit coordinates are very convenient, they are not the basis of euclidean geometry and therefore IMHO it should be beneficial to investigate distance-geometric algorithms and their inherent complexity.  

A further argument in favor of distance based geometric definitions is that they can be used directly in more general settings like on manifolds or discrete structures.


That said, I am astonished that that aspect of my question apparently hasn't played a role in the replies I got so far.



Checking planar convexity of 4 points with Stewart's formula
The given expression is


$$\begin{multline\*}(a' + b')((a + c')(a - c') - a'b') + b'(b + a)(b − a) \\
= a'(a^2 - b'^2 - c'^2) + b'(b^2 - c'^2 - a'^2) = -2a'b'c'(\cos ∠BDC + \cos ∠CDA).\end{multline\*}$$


If the configuration is convex, the intersecting segments must be $AB, CD$ by the condition on $c + c'$, so $∠BDC + ∠CDA < 180^\circ$ and the expression is negative. If the configuration is concave, the outer triangle must be $ABC$ by the condition on $a + b + c$, so $∠BDC + ∠CDA > 180^\circ$ and the expression is positive.




---


Note that the area-based check is also easy to phrase as a stable inequality. If we sort the four areas so that $K\_1 \le K\_2 \le K\_3 \le K\_4$, we need only distinguish between $K\_1 + K\_2 + K\_3 = K\_4$ and $K\_2 + K\_3 = K\_1 + K\_4$, which can be done by checking whether $K\_2 + K\_3 > K\_4$.


