Giải thích các bước giải:
a.Gọi D là trung điểm BC, $BH//B'C', CI//B'C', H, I\in AD$
$\to BH//CI\to\dfrac{BH}{CI}=\dfrac{HD}{DI}=\dfrac{BD}{CD}=1$
$\to BH=CI, DH=DI$
Lại có :
$B'C'//BH, CI$
$\to \dfrac{AB}{AC'}+\dfrac{AC}{AB'}=\dfrac{AH}{AG}+\dfrac{AI}{AG}$
$\to \dfrac{AB}{AC'}+\dfrac{AC}{AB'}=\dfrac{AH+AI}{AG}$
$\to \dfrac{AB}{AC'}+\dfrac{AC}{AB'}=\dfrac{AD-HD+AD+DI}{AG}$
$\to \dfrac{AB}{AC'}+\dfrac{AC}{AB'}=\dfrac{2AD}{AG}$
$\to \dfrac{AB}{AC'}+\dfrac{AC}{AB'}=3$ vì G là trọng tâm $\Delta ABC$
b.Ta có : $C'A'//CI//BH$
$\dfrac{GA'}{CI}=\dfrac{GD}{DI}$
$\to\dfrac{1}{GA'}=\dfrac{DI}{CI.GD}$
$\dfrac{GB'}{CI}=\dfrac{AG}{AI}\to\dfrac{1}{GB'}=\dfrac{AI}{AG.CI}$
$\to\dfrac{1}{GA'}+\dfrac{1}{GB'}=\dfrac{DI}{CI.GD}+\dfrac{AI}{AG.CI}$
$\to\dfrac{1}{GA'}+\dfrac{1}{GB'}=\dfrac{1}{CI}(\dfrac{DI}{GD}+\dfrac{AI}{AG})$
$\to\dfrac{1}{GA'}+\dfrac{1}{GB'}=\dfrac{1}{CI}(\dfrac{DI}{\dfrac 12AG}+\dfrac{AI}{AG})$
$\to\dfrac{1}{GA'}+\dfrac{1}{GB'}=\dfrac{1}{CI}(\dfrac{2DI}{AG}+\dfrac{AI}{AG})$
$\to\dfrac{1}{GA'}+\dfrac{1}{GB'}=\dfrac{1}{CI}(\dfrac{2DI+AI}{AG})$
$\to\dfrac{1}{GA'}+\dfrac{1}{GB'}=\dfrac{1}{CI}(\dfrac{IH+AI}{AG})$
$\to\dfrac{1}{GA'}+\dfrac{1}{GB'}=\dfrac{1}{CI}.\dfrac{AH}{AG}$
$\to\dfrac{1}{GA'}+\dfrac{1}{GB'}=\dfrac{1}{CI}.\dfrac{BH}{CG'}$
$\to\dfrac{1}{GA'}+\dfrac{1}{GB'}=\dfrac{1}{CG'}$ vì $BH=CI$
