设BD分别交AG和AF于P和Q,BE分别交AG和AF于P和Q
如图,△ABC被分为9个部分,设△ABD里的三个部分的面积从左到右分别为a、b、c,△BDE里的三个部分的面积从左到右分别为d、e、f,△BCE里的三个部分的面积从左到右分别为g、h、i,阴影部分的面积就是h.
显然有
a+b+c=1/3
d+e+f=1/3
g+h+i=1/3
a+d+g=1/3
b+e+h=1/3
c+f+i=1/3
AG交△BCD于A、M、G,由梅涅劳斯定理,
BM/MD * DA/AC * CG/GB = 1
BM/MD * 1/3 * 2/1 = 1
BM/MD=3/2
a/(b+c)=BM/MD=3/2
又a+b+c=1/3,所以b+c=1/3 - a
所以a/(1/3 - a)=3/2
a=1/5,
AF交△BCD于A、N、F,由梅涅劳斯定理,
BN/ND * DA/AC * CF/FB = 1
BN/ND * 1/3 * 1/2 = 1
BN/ND=6
(a+b)/c=6
(1/3 - c) / c = 6
c=1/21
所以,
b = 1/3 - a - c = 1/3 - 1/5 - 1/21 = 3/35
AG交△BCE于A、P、G,由梅涅劳斯定理,
BP/PE * EA/AC * CG/GB = 1
BP/PE * 2/3 * 2/1 = 1
BP/PE = 3/4
(a+d)/(b+c+e+f)=3/4
(a+d)/(1/3 - a + 1/3 - d) = 3/4
a+d=2/7
d = 2/7 - 1/5 = 3/35
AF交△BCE于A、Q、F,由梅涅劳斯定理,
BQ/QE * EA/AC * CF/FB = 1
BQ/QE * 2/3 * 1/2 = 1
BQ/QE = 3
(a+b+d+e)/(c+f) = 3
(1/3 - c + 1/3 - f)/(c+f) = 3
c+f=1/6
f = 1/6 - 1/21 = 5/42
所以,
e = 1/3 - d - f = 1/3 - 3/35 - 5/42 = 9/70
所以,
阴影部分的面积h
= 1/3 - b - e
= 1/3 - 3/35 - 9/70
= 5/42