Проведем через точку M прямые, параллельные сторонам
треугольника, и обозначим точки пересечения этих прямых со
сторонами треугольника, как показано на рисунке. Тогда K1 , K2 , K3
-- середины отрезков
A1 A2 ,
B1 B2 ,
C1 C2 соответственно. Поэтому
M K 1 → \displaystyle \overrightarrow{MK_{1}} M K 1 = 1 2 \displaystyle {\textstyle\frac{1}{2}} 2 1 (M A 1 → \displaystyle \overrightarrow{MA_{1}} M A 1 + M A 2 → \displaystyle \overrightarrow{MA_{2}} M A 2 ), M K 2 → \displaystyle \overrightarrow{MK_{2}} M K 2 = 1 2 \displaystyle {\textstyle\frac{1}{2}} 2 1 (M B 1 → \displaystyle \overrightarrow{MB_{1}} M B 1 + M B 2 → \displaystyle \overrightarrow{MB_{2}} M B 2 ), M K 3 → \displaystyle \overrightarrow{MK_{3}} M K 3 = 1 2 \displaystyle {\textstyle\frac{1}{2}} 2 1 (M C 1 → \displaystyle \overrightarrow{MC_{1}} M C 1 + M C 2 → \displaystyle \overrightarrow{MC_{2}} M C 2 ).
Кроме того,
M C 1 → \displaystyle \overrightarrow{MC_{1}} M C 1 + M B 2 → \displaystyle \overrightarrow{MB_{2}} M B 2 = M A → \displaystyle \overrightarrow{MA} M A , M C 2 → \displaystyle \overrightarrow{MC_{2}} M C 2 + M A 1 → \displaystyle \overrightarrow{MA_{1}} M A 1 = M B → \displaystyle \overrightarrow{MB} MB , M A 2 → \displaystyle \overrightarrow{MA_{2}} M A 2 + M B 1 → \displaystyle \overrightarrow{MB_{1}} M B 1 = M C → \displaystyle \overrightarrow{MC} MC .
Тогда
M K 1 → \displaystyle \overrightarrow{MK_{1}} M K 1 + M K 2 → \displaystyle \overrightarrow{MK_{2}} M K 2 + M K 3 → \displaystyle \overrightarrow{MK_{3}} M K 3 = 1 2 \displaystyle {\textstyle\frac{1}{2}} 2 1 (M A 1 → \displaystyle \overrightarrow{MA_{1}} M A 1 + M A 2 → \displaystyle \overrightarrow{MA_{2}} M A 2 + M B 1 → \displaystyle \overrightarrow{MB_{1}} M B 1 + M B 2 → \displaystyle \overrightarrow{MB_{2}} M B 2 + M C 1 → \displaystyle \overrightarrow{MC_{1}} M C 1 + M C 2 → \displaystyle \overrightarrow{MC_{2}} M C 2 ) =
= 1 2 \displaystyle {\textstyle\frac{1}{2}} 2 1 $\displaystyle \left(\vphantom{(\overrightarrow{MC_{1}} + \overrightarrow{MB_{2}...
...htarrow{MA_{1}})
+ (\overrightarrow{MA_{2}} + \overrightarrow{MB_{1}})}\right.( ( ( \displaystyle \overrightarrow{MC_{1}}+ + + \displaystyle \overrightarrow{MB_{2}}) + ( ) + ( ) + ( \displaystyle \overrightarrow{MC_{2}}+ + + \displaystyle \overrightarrow{MA_{1}}) + ( ) + ( ) + ( \displaystyle \overrightarrow{MA_{2}}+ + + \displaystyle \overrightarrow{MB_{1}}) ) ) \displaystyle \left.\vphantom{(\overrightarrow{MC_{1}} + \overrightarrow{MB_{2}...
...htarrow{MA_{1}})
+ (\overrightarrow{MA_{2}} + \overrightarrow{MB_{1}})}\right)$ =
= 1 2 \displaystyle {\textstyle\frac{1}{2}} 2 1 (M A → \displaystyle \overrightarrow{MA} M A + M B → \displaystyle \overrightarrow{MB} MB + M C → \displaystyle \overrightarrow{MC} MC ) = 1 2 \displaystyle {\textstyle\frac{1}{2}} 2 1 . 3 . M O → \displaystyle \overrightarrow{MO} MO = 3 2 \displaystyle {\textstyle\frac{3}{2}} 2 3 M O → \displaystyle \overrightarrow{MO} MO .