The area of triangle ABC is 120 cm2, e is the midpoint of AC, and f is the bisector of BC. Find the area of the shaded part.
Parallel lines intersect with D as AC, and intersect with Be and AB at points M and N respectively, then △GMD∽△GEA, and DN=2AC/3, DM=DN/2=AC/3, AE=AC/2, so DM: AE = 2: 3, and the area of △ BCE is1. GM = 2em/5 = 2be/ 15 and MB = 2be/3, so GM:MB= 1:5, so the area of △GMD is 1/5 of the area of △GBD, that is, 80/3×1. That is, 16/3×9/4= 12 (square centimeter), so the area of the shaded part is 80/3+ 16/3+ 12=44 (square centimeter). ?