Proving that $\cos(\pi-\phi)=-\cos\phi$ geometrically I want to geometrically prove that $\cos(\pi-\phi)=-\cos\phi$ without resorting to the unit circle or trigonometric formulas, but have difficulties figuring it out.
It's easy enough to do the sine, however: you draw a right triangle to complement the existing scalene triangle and then subtract the area of the smaller right triangle from the bigger one (see picture).

$$a=c\sin\phi$$
$$S_x=\frac{c \cdot \sin\phi \cdot (b+b')}{2}-\frac{c \cdot \sin\phi \cdot b}{2} = \frac{c \cdot \sin\phi \cdot b'}{2}$$
On the other hand, $$S_x=\frac{c \cdot \sin(\pi-\phi) \cdot b'}{2}$$
Thus, $$ \sin(\pi-\phi) = \sin \phi$$
I'll be grateful for any ideas/advice.

Thanks to @Blue for the idea; also, please check out the post referenced in the question's comments, since there may be a fundamental flaw in this approach to reasoning altogether.
Still, let us build an isosceles scalene triangle ($c=b'$). In that case, the longer hypotenuse can be calculated twofold:
$$(b+b')^2+(b'sin\phi)^2=2b'^2 - 2b'^2cos(\pi-\phi)$$
Knowing that $b=b'cos\phi$,
$$b^2+2bb'+b'^2+b'^2sin^2\phi=2b'^2-2b'^2cos(\pi-\phi)$$
$$b'^2 cos^2\phi + b'^2 sin^2\phi + 2b'^2cos\phi=b'^2-2b'^2cos(\pi-\phi)$$
$$cos\phi=-cos(\pi-\phi)$$
