Giải thích các bước giải:

a.Xét $\Delta ABH,\Delta ABC$ có:

Chung $\hat B$

$\widehat{AHB}=\widehat{BAC}(=90^o)$

$\to \Delta HBA\sim\Delta ABC(g.g)$

$\to \dfrac{AB}{BC}=\dfrac{HB}{AB}$

b.Vì $\Delta ABC$ vuông tại $A$

$\to BC^2=AB^2+AC^2$

$\to BC=\sqrt{AB^2+AC^2}$
$\to BC=\sqrt{18^2+24^2}$

$\to BC=30$

vì $CD$ là phân giác $\hat C$

$\to \dfrac{DA}{DB}=\dfrac{CA}{CB}=\dfrac{24}{30}=\dfrac45$

$\to \dfrac{DA}4=\dfrac{DC}5=\dfrac{DA+DC}{4+5}=\dfrac{AB}9=\dfrac{18}9=2$

$\to AD=8, DC=10$

c.Xét $\Delta BHF,\Delta BEC$ có:

Chung $\hat B$

$\widehat{BHF}=\widehat{BEC}(=90^o)$

$\to \Delta BHF\sim\Delta BEC(g.g)$

$\to \dfrac{BH}{BE}=\dfrac{BF}{BC}$

$\to BH.BC=BE.BF$

Mà $BH.BC=BA^2$

       $BA=BG$

$\to BE.BF=BG^2$

$\to \dfrac{BE}{BG}=\dfrac{BG}{BF}$

$\to \Delta BEG\sim\Delta BGF(c.g.c)$
$\to \widehat{BGF}=\widehat{BEG}=90^o$

$\to BG\perp GF$