Square in Arbelos without inversion

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.

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.