Các tam giác $ABC$ và $A’B’C’$ có trọng tâm lần lượt là $G$ và $G’$. chứng minh rằng:
                                 $\overrightarrow {GG'}  = \frac{1}{3}\left( {\overrightarrow {AA'}  + \overrightarrow {BB'}  + \overrightarrow {CC'} } \right)$

Ta có :
$\left\{ \begin{array}{l}
\overrightarrow {GG'}  = \overrightarrow {GA}  + \overrightarrow {AA'}  + \overrightarrow {A'G'} \\
\overrightarrow {GG'}  = \overrightarrow {GB}  + \overrightarrow {BB'}  + \overrightarrow {B'G'} \\
\overrightarrow {GG'}  = \overrightarrow {GC}  + \overrightarrow {CC'}  + \overrightarrow {C'G'}
\end{array} \right.$
Cộng từng vế ba đẳng thức trên ta có :
$3\overrightarrow {GG'}  = \left( {\overrightarrow {GA}  + \overrightarrow {GB}  + \overrightarrow {GC} } \right) + \left( {\overrightarrow {AA'}  + \overrightarrow {BB'}  + \overrightarrow {CC'} } \right) + \left( {\overrightarrow {A'G'}  + \overrightarrow {B'G'}  + \overrightarrow {C'G'} } \right) $
$= \overrightarrow {AA'}  + \overrightarrow {BB'}  + \overrightarrow {CC'} $
Vậy $\overrightarrow {GG'}  = \frac{1}{3}\left( {\overrightarrow {AA'}  + \overrightarrow {BB'}  + \overrightarrow {CC'} } \right)$

Hệ quả : hai tam giác $ABC$ và $A’B’C’$ có cùng trọng tâm khi và chỉ khi: $\overrightarrow {AA'}  + \overrightarrow {BB'}  + \overrightarrow {CC'}  = \overrightarrow 0 $.

minh hoc toan yeu lam.neu ban co bai giai tren nay.xin ban vui long chu thich sau cau giai cua tung cau tren bai toan la dinh nghia cua bai toan di kem de minh on lai kien thuc co ban.cam on ban –  phuctoanngo 10-10-12 06:10 AM

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