<Qua-2025-790>, hint

Let AB=a, AD=b, b>a  r',O' are the radius and center of s', similarly r", O" are those  of  s"

angle bisectors of <A, <C cut the BD at A', C' respectively;

<BAC=<DCA=x, <DAC=y=<BCA,

<CAO'=<ACO"=x-(x+y)/2=(x-y)/2, BA'=DC'=BD a/(a+b). Hence A'C'=BD-(BA'+DC')=

BD (b-a)/(a+b), which is constant

AO'=r'/sin(A/2), CO"=r"/(sinA/2). <KAO'=<KCO", <AKO'=<CKO" Triangles KAO' and KCO" are similar

.O'K(=r')/AK=O"K(=r")/CK=k, Hence O'O"= r'+r"=k(AK+CK)=kAC ,constant

Since A'C', O'O" are both constant O'O" are parallel to each other