<Cir-2023-749>, hint

By angle chasing,  <APH=105, < PAH=30,  <PHA=45

AH/sin<APH=AH/sin105=R'=AH/sin75=AH/sin<ADH=R.

R: radius of circumcircle of he dodecagon, R'; radius of circle S

O: center of the circumcircle of the dodecagon, O' : the center of circle S.

Since AH is the intersection of two equal circles, O and O' are symmetric about AH.

Since triangles GAH and AHN are congruent, <AHG=90, GH//OO'. O' lies on NH.

O'P=OE=R'=R, <EOG=60=<PO'G=2<PAH(=30), OO'PE is parallelogram and  so PE=OO'

Triangles OHO' and OGH are congruent. OO'=GH=PE.