Just another provocative triangle
Consider a triangle ABC with AB=3, BC=6, CA=5. Let the angle bisector of angle BAC intersect side BC at a point D and E is the middle of the segment BD. If I is the incenter of the triangle ABC and F∈(AC) with FC=k FA , find the INTEGER value of k so that E,I,F are collinear. [k]=?
I.E. The integer value of 25.623=[25.623]=25 ; [0.124]=0 .