Square in Arbelos without inversion I am looking for a way to prove that the $NPGH$ polygon is a square without using the Pappus' chain theorem (without inversion). I know that $NPGH$ is a rectangle so the question is to prove that $GH=HP$.
I know $AN/NP=NP/BP$ , $MP=GL$ and $HM=LN$. $BP.BC=BM.BK=...=BN.BA$ etc. but I can't conclude.
(The arbelos here is a special case because $B$ divides $AC$ according to the golden ratio).

 A: The result comes as a by-product of the determination of the radius of the inner circle. Let $r_1=AC$ and $r_2=BD$ be the radii of the inner half-circles (see figure below), $AO=r_1+r_2$ the radius of the outer half-circle, $r=GP$ the radius of the inscribed circle. Set in addition: $h=GH$, $x=DH$, implying $CH=r_1+r_2-x$.
Pythagoras' theorem, applied to triangles $GHD$, $GHC$, $GHO$, gives:
$$
h^2=(r_2+r)^2-x^2=(r_1+r)^2-(r_1+r_2-x)^2=(r_1+r_2-r)^2-(r_1-x)^2.
$$
These are two equations for $x$ and $r$, which can be solved giving:
$$
r={r_1r_2(r_1+r_2)\over r_1^2+r_1r_2+r_2^2},
\quad
x={r_2^2(2r_1+r_2)\over r_1^2+r_1r_2+r_2^2}.
$$
Inserting these into one of the equations for $h$, one then finds $h=2r$, as required.

A: I finally found another solution :
a) Using proposition 1 of the book of lemmas of Archimedes, we know that $AGD$, $CHD$, $AEH$, $CFG$, $BEG$ and $BFH$ are straight lines.
b) It is easy to show that the lines $BK$ and $AD$ are parallel as are the lines $BI$ and $CD$.
Therefore $AB:BC = AL:LH=AN:NP$, and $BC:AC=CM:MG=CP:PN$. Hence $AN:NP=NP:CP$ (1).
(All this is used in proposition 6 of the book of lemmas.)
c) Now we can notice that the right triangles $AGN$ and $PCH$ are similar. Thus $GN:AN=CP:HP=CP:GN$ (2).
d) With (1) and (2) $GN^2=NP^2$ so $GN=NP$. The rectangle NPHG is a square.
