Từ $O$ kẻ $OK\perp BC$
Từ $A$ kẻ $AH\perp BC$
$\Rightarrow AH//OK$
$\Rightarrow \dfrac{OK}{AH} = \dfrac{OA'}{AA'}$ (Theo định lý $Thales$ đảo)
$\Rightarrow \dfrac{\dfrac{1}{2}OK.BC}{\dfrac{1}{2}AH.BC} = \dfrac{OA'}{AA'}$
$\Rightarrow \dfrac{S_{BOC}}{S_{ABC}} = \dfrac{OA'}{AA'}$
Chứng minh tương tự, ta được:
$\dfrac{S_{AOC}}{S_{ABC}} = \dfrac{OB'}{BB'}$
$\dfrac{S_{AOB}}{S_{ABC}} = \dfrac{OC'}{CC'}$
Ta được:
$\dfrac{OA'}{AA'} + \dfrac{OB'}{BB'} + \dfrac{OC'}{CC'} = \dfrac{S_{BOC}}{S_{ABC}} + \dfrac{S_{AOC}}{S_{ABC}} + \dfrac{S_{AOB}}{S_{ABC}} = \dfrac{S_{ABC}}{S_{ABC}} = 1$
