A regular Polygon of an even number of sides being inscribed in a circle as ABC...A'...C'B'A' so that AA' is a diameter if two angular points next but one to each other as B, B' be joined, and the other lines parallel to BB' and joining pairs of angular points be drawn as CC', DD',...., then

(BB' + CC' +.... ) / AA' = A'B / BA

Since the two angles are equal in two traingle ABF and B'FK then the third angle is equal which are
<k = <A so these two triangle are congruent .

so BF/FA = B'F/FK