giúp mình hết bài này nhé

@nan 



$\text{a)}$ 

$\text{Ta có $\overline{MA}$ + $\overline{2MB}$ + $\overline{MC}$ + $\overline{MD}$ = $\overline{3MO}$ }$

  $\text{⇔ ( $\overline{MO}$  + $\overline{OA}$  ) + ( $\overline{MO}$  + $\overline{OB}$  ) +  ( $\overline{MO}$ + $\overline{OC}$ ) + ( $\overline{MO}$ + $\overline{OD}$ ) + $\overline{MB}$ - $\overline{3MO}$  = $\overline{O}$ }$

  $\text{⇔ ( $\overline{OA}$ + $\overline{OC}$ )  + ( $\overline{OB}$ + $\overline{OC}$ ) + ( $\overline{MO}$ + $\overline{MB}$ ) = $\overline{0}$ ⇔ $\overline{0}$ + $\overline{0}$ + ( $\overline{MO}$ + $\overline{MB}$ ) = $\overline{0}$ }$  

$\text{⇔ $\overline{MO}$ + $\overline{MB}$ = $\overline{0}$ }$  

$\text{Suy ra M là trung điểm của đoạn thẳng OB }$ 
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$\text{b)}$ 

$\text{ta có }$ 

$\text{$\overline{3AN}$ = $\overline{AB}$ + $\overline{AC}$ + $\overline{AD}$ ⇔ $\overline{3AN}$ = ( $\overline{AB}$ + $\overline{AD}$ ) + $\overline{AC}$}$ 

$\text{⇔ $\overline{3AN}$ = $\overline{AC}$ + $\overline{AC}$ ⇔ $\overline{AN}$ = $\frac{2}{3}$  $\overline{AC}$ }$ 

$\text{Suy ra N thuộc đoạn thẳng AC sao cho AN = $\frac{2}{3}$  AC }$ 
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$\text{c)}$  

$\text{ta có }$ 

$\text{$\overline{IA}$ + $\overline{IB}$ + $\overline{IC}$ = $\overline{4ID}$ }$ 

$\text{⇔ ( $\overline{IO}$ + $\overline{OA}$ ) + ($\overline{IO}$ + $\overline{OB}$ ) + ( $\overline{IO}$ +  $\overline{OC}$ ) - 4 ( $\overline{IO}$ + $\overline{OD}$ ) = $\overline{0}$ }$ 
 
$\text{⇔ ( $\overline{OA}$ + $\overline{OC}$ ) + $\overline{OB}$ - $\overline{IO}$ - $\overline{4OD}$ =$\overline{0}$ ⇔  $\overline{0}$ + $\overline{OB}$ + $\overline{IO}$ - $\overline{4BO}$  =  $\overline{0}$  ⇔ $\overline{5OB}$   + $\overline{OI}$   = $\overline{0}$   }$ 

$\text{⇔  $\overline{OI}$   = $\overline{5BO}$   }$ 

$\text{suy ra i thuộc đường thẳng BD sao cho O nằm giữa B , I VÀ OI = 5BO }$