This is a well known math problem also called the Feynman's triangle. If the ABC is equilateral triangle, such that the bd/bc=1/3 and ce/ca=1/3, af/ab=1/3. Then the ratio of the area of the shaded region to area of abc is 60.
This is the work to the problem;
BE2=CE2+CB2−2.CE.CBcos(60) BE2=12+32−2.1.312 BE=7√ (1)