How can we visualize $a^2+b^2-c^2$ for a triangle of sides $a$, $b$, $c$? Let $a$, $b$, and $c$ be three lengths of sides of a triangle, that is, $a+b>c$.
How can we visualize the value $a^2+b^2-c^2$ as length of some segment or area, ... constructed from the triangle $ABC$ ?
The value appears in some context, for example Law of Cosine, barycentric coordinate of orthocenter.
I searched, and tried myself for a while, but still have no clues.

 A: 
The law of cosines is:
$$a^2+b^2=c^2+2ab\cos C$$
$ab\cos C$ is the dot product of the vectors $\vec a$ and $\vec b$.
It therefore says that:
$$c^2=(a')^2+(b')^2$$
A: Here is one way of doing a construction to illustrate the area $a^2+b^2-c^2$:
From the Cosine Rule we know this quantity is equivalent to $2ab\cos C$.
Therefore at $C$ draw a line $CD$ perpendicular to $CA$ and of length $b$, so that $\angle BCD=90-C$.
Now draw a perpendicular from $D$ onto $BC$ (which has length $a$), meeting $BC$ at $E$. The line $DE$ has length $b\cos C$.
Finally you can construct a rectangle whose base is BC and height is $2\times DE$, and this rectangle has the required area.
A: It represents work or energy experienced rather than what can be seen as a pure geometric quantity.
From Cosine Rule
$$ a^2+b^2-c^2 = 2ab \cos C \;$$
This is the work done by force a when displaced through distance 2 b.  For example no work done by a person walking on frictionless ground as the vectors are perpendicular and dot product is zero.

A: 
Here is one possible geometric construction:

*

*Draw $Circle1$ with diameter $AC$.

*Draw $Circle2$ with center $A$ and radius $c$. Let $Circle2$ intersect $Circle1$ at $D$

*Draw $Circle3$ with center $D$ and radius $a$.

*Extend $AD$ to meet $Circle3$ at $E$.

*Construct square $CEFG$.

As an exercise, calculate the area of the square $CEFG$.
