In acute △ABC with circumcircle Γ and incentre I, the incircle touches side AB at F. The external angle bisector of ∠ACB meets ray AB at L. Point K lies on the arc CB of Γ not containing A, such that ∠CKI=∠IKL. Ray KI meets Γ again at D=K. Prove that ∠ACF=∠DCB. geometrycircumcircleincenterangle bisector