If ABCDEF is cyclic, then (intersection of AB,DE), (intersection of BC,EF), (intersection of CD,FA) are collinear

(if A = B, for instance, "AB" is the tangent to A = B)

=>

If ABCD is cyclic, then (intersection of tangentA, tangentC), (intersection of AB,CD), (intersection of BC,AD), (intersection of tangentB,tangentD) are collinear

+

angle chase: (center of circle), A, C, (intersection of tangentA, tangentC) are cyclic, with the first and last points diametrically opposite
(center of circle), B, D, (intersection of tangentB, tangentD) are cyclic, with the first and last points diametrically opposite

=>

(center of circle) is on radical axis of above two circles;

;;

(intersection of AC,BD) := P also satisfies PA*PC = PB*PD
hence is also on radical axis of above two circles;

=> AP is perpendicular to connection of centers of above two circles

by homothecy about (center of circle), AP is perpendicular to connection of (intersection of tangentA, tangentC), (intersection of tangentB, tangentD), aka (intersection of AB, CD), (intersection of BC, AD)