ABC三角型 中點 比例問題

***Using vector to solve this problem***
for clarity, all sides mentioned below are vectors.


let BC= a(vector) , BA=b(vector)
ratio of |BE|:|EC| = t : 1-t (t is scalar)

BD = a/2+b/2
BF = (a+b)/4

AC = a - b
AD = (a-b)/2

AE = BE - BA = ta - b
AF = AB/2+AD/2 = -b/2 + (a-b)/4 = a/4 - 3b/4

since directional vector of AF = directional vector of AE
so 3/4 = 1/4t
=> t = 1/3
=> 1-t = 2/3
=>|BE|:|EC| = t : 1-t = 1:2

2013-01-17 23:02:19 補充：
corrected one:

BD = a/2+b/2
BF = (a+b)/4

AE = BE - BA = ta - b
AF = BF - BA = (a+b)/4 - b = a/4 - 3b/4

since directional vector of AF = directional vector of AE
so (-3/4)/(1/4) = (-1)/t
=> t = 1/3
=> 1-t = 2/3
=>|BE|:|EC| = t : 1-t = 1:2

No, haven't learnt.

Find a point P on BC such that DP // AFE, by mid-pt theorem,
as BF = FD, so, BE = EP, and
AD = DC, so, EP = PC.
Therefore, BE = EP = PC
ie. BE : EC = 1 : 2

Have you ever heard of Menelauss Theorem?