(O), (P) are circles with diameters AC > AB tangent at A. Perpendicular bisector of BC cuts (O) at E, F, so that triangle BEC is isosceles (or BECF is a rhombus). Circle (Q) is internally / externally tangent to (O), (P) and to BE. Prove BQ is perpendicular to AC.

Proof with no calculations:

EB, FB cut (O) again at K, L. FE bisects angle <CFL = <CFB => EL = EC = EB => B is incenter of triangle CKL. Perpendicular to AC at B cuts (O) at M, N. (O) cuts circle (B) with center B and diameter MN in diametrically opposite points => inversion in (B) with negative power -BM^2 takes (O) into itself. Line EBK through the inversion center also goes to itself. Power of B to (O) is -IM^2 = IM * IN = IA * IC => the inversion takes A to C. Circle (P) with diameter AB goes to line perpendicular to AC at C, the external bisector of the angle <KCL. Let J_k, J_l be K-, L-excenters of triangle CKL. The inversion image (Q') of (Q) is tangent to J_KJ_l, to BE = BJ_k, and externally tangent to the 9-point circle (O) of the triangle BJ_kJ_l, therefore (Q') is its excircle in the angle <BJ_kJ_l, centered on the external bisector of the angle <EBF = J_kBJ_l, perpendicular to the internal bisector BC of this angle, therefore BQQ' is perpendicular to AC.

This permits generalization to a non-isosceles triangle CKL with circumcircle (O) and incenter B and angle bisector CBA. (P) is circle with diameter AB, (Q) is circle internally tangent to (O), externally tangent to (P), and to the angle bisector KBE. No change in the proof that (Q) is also tangent to the other angle bisector LBF.

https://www.cut-the-knot.org/Curriculum/Geometry/InversionInCircleAndIsosceles2.shtml

BTW, for the sake of credits, Are you "the Vladimir"?