✔ 最佳答案
a. Since D is the mid-point of BC, thus d=(b+c)/2
Similarly, we have e=(a+c)/2 and f=(a+b)/2
Thus
d+e+f = (b+c)/2 + (a+c)/2 + (a+b)/2 = a+b+c
b. Since G is the centroid, we know that AG:GD=2:1
Thus
g = (a+2d)/(1+2) = (a+2d)/3 = (a+b+c)/3
By (a), a+b+c = d+e+f,
thus
g = (a+b+c)/3 = (d+e+f)/3 = i
Remark. This result means if you consider ΔDEF
as a "mid-point triangle" of ΔABC,
then they actually have the same centroid.
c. Using the same argument in (a), we have
d = (b+c)/2
e = (a+c)/2
f = (a+b)/2
Thus
b+c = 2d = -2i+6j.........(1)
a+c = 2e = 2i-2j............(2)
a+b = 2f = 6i-4j............(3)
a+b+c = 3g = 3i............(4)
(4)-(1): a = 5i-6j
(4)-(2): b = i+2j
(4)-(3): c = -3i+4j
