<tRI-2025-2267>, hint

A line through B, parallel to AK cut the line CA at A'. Since DK=DA, DA'=DB (BK+AD)

D'C=DC, D' on line AC, then triangles ACD and ACD' are congruent, <AD'D=<ADD'=2<B

2<B=<KAD+>AKD=2<ABC. Suppose the bisector of <AD'B cut AB perpendicularly

. Thus AD'=BD'. Finally  BD(BK+AD)=BD'=D'A(2CD+AD). Hence BK=2CD.