Zoran Skoda Ptolemy from Pluecker

If $a$ is a chord in a circle of radius $r$ and its inscribed angle is $\alpha$ then $a = 2 r sin\alpha$. If $(x_1,y_1)$ and $(x_2,y_2)$ are the coordinates of the ends $1$ and $2$ of the chord with the origin in the center of the circle, then the corresponding determinant $D_{1 2}$ is, up to the sign, the area of the paralelogram spanned by the radii-vectors is $r^2 sin(2\alpha)$. So if we take four points $1,2,3,4$ with chords $a = \bar{12},b=\bar{23},c=\bar{34},d=\bar{41}$ in the counterclockwise direction and the corresponding incidence angles $\alpha,\beta,\gamma,\delta$ the Ptolomy theorem $e f = a c + b d$ (where $e = \bar{13}$ and $f=\bar{24}$ is up to rescaling by $2 r$,

$$sin(\alpha+\beta) sin(\beta+\gamma) = sin(\alpha)sin(\gamma)+sin(\beta)sin(\delta).$$

We obtain this identity from Pluecker identity for ANOTHER chordal quadrangle. Namely, we make new chords $a_1,b_1,c_1$ shorter than $a,b,c$ so that the corresponding incidence angles are $\alpha_1 = \alpha/2$, $\beta_1 = \beta/2$ and $\gamma_1 = \gamma/2$; they add up to less than $\pi/2$, hence all 4 chords are in the same semicircle. That means that the incidence angle $\delta_1 = \pi-(\alpha+\beta+\gamma)/2$ on the fourth chord $d_1$ is viewed from the center from outside, hence in the clockwise direction when going from vertex $4$ to vertex $1$, thus the determinant $D_{41}=-D_{14}$ for that quadrangle is negative times the area! All the others in the Pluecker relation

$$D_{13}D_{24} = D_{12}D_{34}+D_{23}D_{14}$$

are clearly positive in our conventions (regardless the position and shape of the original chordal rectangle with respect to the center). Clearly, $D_{12} = r^2 sin(\alpha) = r a/2$, $D_{13} = r e/2$, and so on, but $D_{41} = -r^2 |sin(2\delta_1)|$. Now, $sin(2\delta_1) = sin(2\pi-(\alpha+ \beta+\gamma))$, this is $sin(2\pi-(\alpha+\beta+\gamma)) = sin(\pi+\delta) = - sin(\delta)$ (Note that $sin(\pi+\delta)$ is negative itself as $2\pi\gt \pi+\delta\gt\pi$). Thus, $D_{41} = -r^2 sin(\delta) = -r d/2$ and $D_{14} = r d/2$. After multiplying all by $2/r$ we obtain $e f = a c + b d$.