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 .
The answer is 2.
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