Problem: In the figure, CD,AE and BF are one-third of their respective sides. It follows that AN2​:N2​N1​:N1​D=3:3:1, and similarly for lines BE and CF. Then the area of triangle N1​N2​N3​ is:
Answer Choices:
A. 101​△ABC
B. 91​△ABC
C. 71​△ABC
D. 61​△ABC
E. none of these
Solution:
By subtracting from △ABC the sum of △CBF,△BAE, and △ACD and restoring △CDN1​+△BFN3​+△AEN2​, we have △N1​N2​N3​.
△CBF=△BAE=△ACD=31​△ABC.
From the assertion made in the statement of the problem, it follows that △CDN1​=△BFN3​=△AEN2​=71​⋅31​△ABC=211​△△ABC.
∴△N2​N2​N3​=△ABC−3⋅31​△ABC+3⋅211​△ABC=71​△ABC.
