<Cir-2024-759>, hintL

Let's prove triangles APC and FBE are similar . Then lines AF, BP, EC are concurent

Angle chasing ; Let N be the intersection of circle #1 and PF ( O . center of circle #1)

<CAP=<CBP= <NAB (PB is tangent to circle #1)=<EFB

<PEB=<EFB+<EBF=<PAB=<PBA. Hence <PAN=<NBA=<APC

Triangles ACP and FEB are similar. Hence AF, BP, EC concur at center of homothety.

<APC=<ABC=<PAN( PA tangent). PC//NA, MA=MP Triangles MPC and MAN are congruent

ACPN are parallelogram AC//FP, <CAP=<CBP=<NAB( PA tangent)=<NFB

Hence AP//FB and Triangles ACP and FEB are similar.