A geometrical calculation in Fresnel's paper "Memoir on the diffraction of light" 1819 It is a geometrical problem which I find difficult to solve reading the Fresnel's paper "Memoir on the diffraction of Light".
According to the figure Fresnel sets $z$ as the distance of the element $nn'$ from the point $M$---- (I suppose $z=nM$)-----, $a=CA$, $b=AB$, $IMA$ is an arc with center $C$, $EMF$ is an arc with center $P$ tangential to the point $M$ with the first arc. Eventually, Fresnel calculate that the distance $$nS=\frac{z^2(a+b)}{2ab}.$$ ( I believe that it is an approximation saying $nS≈\frac{z^2(a+b)}{2ab}$) 
(1) How does he find that result?
(2) In my attempt I find that $nS≈\frac{z^2}{2PM}$, pretty close but I cannot find the $PM$ value.Are there any ideas?
(3) Υοu can find the original paper here (page 119): https://archive.org/stream/wavetheoryofligh00crewrich#page/118

 A: 
First, we have that the triangles $RAB$ and $RTC$ are similar, so that
$$TC=RC\,\left(\frac{AB}{RB}\right)=(a+b)\,\left(\frac{c/2}{a}\right)=\frac{(a+b)c}{2a}\,.$$
This means
$$\begin{align}
RF&=\sqrt{FC^2+RC^2}=\sqrt{(FT+TC)^2+RC^2}=\sqrt{\left(x+\frac{(a+b)c}{2a}\right)^2+(a+b)^2}
\\
&=(a+b)\,\sqrt{1+\left(\frac{x}{a+b}+\frac{c}{2a}\right)^2}\approx(a+b)\left(1+\frac{1}{2}\,\left(\frac{x}{a+b}+\frac{c}{2a}\right)^2\right)
\\
&=a+b+\frac{x^2}{2(a+b)}+\frac{cx}{2a}+\frac{(a+b)c^2}{8a^2}\,.
\end{align}$$
(We expand the square with binomial series and receive the approximation)
Next, 
$$\begin{align}
RA&=\sqrt{RB^2+AB^2}=\sqrt{a^2+\left(\frac{c}{2}\right)^2}=a\,\sqrt{1+\left(\frac{c}{2a}\right)^2}
\\&\approx a\,\left(1+\frac{1}{2}\,\left(\frac{c}{2a}\right)^2\right)=a+\frac{c^2}{8a}
\end{align}$$
and
$$\begin{align}
AF&=\sqrt{AM^2+MF^2}=\sqrt{BC^2+(FC-MC)^2}=\sqrt{BC^2+(FC-AB)^2}
\\
&=\sqrt{b^2+\left(x+\frac{(a+b)c}{2a}-\frac{c}{2}\right)^2}=b\,\sqrt{1+\left(\frac{x}{b}+\frac{c}{2a}\right)^2}
\\
&\approx b\,\left(1+\frac{1}{2}\,\left(\frac{x}{b}+\frac{c}{2a}\right)^2\right)
=b+\frac{x^2}{2b}+\frac{cx}{2a}+\frac{bc^2}{8a^2}\,.
\end{align}$$
Finally, we get
$$\begin{align}d&=RA+AF-RF
\\
&\approx\small\left(a+\frac{c^2}{8a}\right)+\left(b+\frac{x^2}{2b}+\frac{cx}{2a}+\frac{bc^2}{8a^2}\right)-\left(a+b+\frac{x^2}{2(a+b)}+\frac{cx}{2a}+\frac{(a+b)c^2}{8a^2}\right)
\\
&=\frac{x^2}{2b}-\frac{x^2}{2(a+b)}=\frac{ax^2}{2b(a+b)}\,.
\end{align}$$
We transform the figure into it's mirror one and take approximatly that $z=m'A$:

Triangles $RTC$, $ARB$ are similar (Fig. 1). So: $$\frac{TC}{AB}=\frac{a+b}{a}$$
Also, triangles $FRC$, $m'RB$ are similar. So:  $$\frac{FC}{m'B}=\frac{a+b}{a}$$
This means that: $(x= FT)$
$$\frac{x}{m'A}=\frac{TC}{AB}=\frac{a+b}{a}$$
With $z≈m'A$ we have:
$$\frac{x}{z}=\frac{a+b}{a}\tag{1}$$
We have calculate above the difference d, as:
$$d=\frac{a}{2b(a+b)}x^2\tag{2}$$
Combined (1) and (2):
$$d= nS =\frac{a+b}{2ab}z^2\tag{3}$$
