Proof of Ptolemy's inequality using $z \mapsto \frac1z$ Ptolemy's inequality: if $A,B,C,D$ are the vertices of a convex quadrilateral $ABCD$, then: $AC \times BD \le AD \times BC + AB \times DC$, with equality only if this quadrilateral is inscribed in a circle. 
There's a certain proof which goes like this:
Consider the transformation $T: z \mapsto \frac1z$. WLOG suppose that $A = 0$, $B = z_1$, $C =z_2$, $D = z_3$. 
$$\left| T(B) - T(D) \right| \le \left| T(B) - T(C)\right| + \left| T(C) - T(D) \right| \implies ... \implies |z_2||z_1 - z_3| \le |z_3||z_1 - z_2| + |z_1| |z_2 - z_3|$$
Which proves the inequality. Then, it says that since $z \mapsto \frac1z$ transforms a circle into a line and vice versa, then equality holds iff $z_1^{-1},z_2^{-1}, z_3^{-1}$ are collinear iff $z_1, z_2, z_3$ are on the same circle. 
I understand why the proof is especially attractive and stuff, but I am confused as why does it also follow that $A$ is on the same circle as well? 
I hope that I'm not missing something too obvious. Thank you. 
 A: You are inverting with center $A$. Inversion maps circles which pass through $A$ to lines not through $A$ and viceversa, and lines through $A$ are sent to lines lines through $A$. After the inversion, $B,C,D$ are on a line, hence before the inversion either they were on a line through $A$ (impossible because we started with a convex polygon) or on a circle through $A$, that is $ABCD$ is cyclic.

Let me answer to the comment. Call $f$ your inversion, $f(z)=\frac{1}{z}$ (usually it is $f(z)=\frac{1}{\bar{z}}$ but the two only differ by a simmetry so never mind). So $$f(x+iy)=\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}.$$ We want to compute the image of any line not through the origin $A=0$. Modulo rotations, we can work on a line of the form $x=a$ with $a\in\mathbb{R}$. For a point on the line we can write $z=a(1+i\tan\theta)$ (you can try to do the same computation with $z=a+iy$, but I find it more coumbersome). The image is $$f(z)=\frac{a}{a^2(1+\tan^2\theta)}-i\frac{a\tan\theta}{a^2(1+\tan^2\theta)}=\frac{1}{a}\left(\cos^2\theta-i\sin\theta\cos\theta\right)=\frac{1}{2a}\left(1-(cos2\theta+i\sin2\theta)\right)$$ which, as $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$, describes a circle of centre $(0, \frac{1}{2a})$. Actually, it is a circle minus a point, which is the origin $A$: to see it, plug in $\theta=\frac{\pi}{2}$ (or $-\frac{\pi}{2}$).
