Giải thích các bước giải:

a.Xét $\Delta ADE,\Delta ABC$ có:

Chung $\hat A$

$\dfrac{AD}{AC}=\dfrac15=\dfrac{AE}{AB}$

$\to\Delta AED\sim\Delta ABC(c.g.c)$ 

b.Ta có: $DE//BH$

$\to \dfrac{AE}{AH}=\dfrac{AD}{AB}=\dfrac4{15}$

$\to AH=\dfrac{15}4AE=\dfrac{45}{4}$

c.Xét $\Delta DEF,\Delta BHC$ có:

$\widehat{DFE}=\widehat{BCH}$ vì $DF//BC$

$\widehat{DEF}=\widehat{BHC}$ vì $DE//BH$

$\to\Delta EDF\sim\Delta HBC(g.g)$

$\to\dfrac{DE}{HB}=\dfrac{DF}{BC}$

$\to\dfrac{DE}{DF}=\dfrac{HB}{BC}$

Ta có $DF//BC$

$\to\dfrac{AD}{AB}=\dfrac{AF}{AC}$

$\to\dfrac{AD}{AF}=\dfrac{AB}{AC}$

Lại có: $\Delta ADE\sim\Delta ABC$

$\to\widehat{ADE}=\widehat{ACB}$

Xét $\Delta ABH,\Delta ABC$ có:

Chung $\hat A$

$\widehat{ABH}=\widehat{ADE}=\widehat{ACB}$

$\to\Delta ABH\sim\Delta ACB(g.g)$

$\to\dfrac{AB}{AC}=\dfrac{BH}{BC}$

$\to\dfrac{AD}{AF}=\dfrac{DE}{DF}$

$\to AD\cdot DF=AF\cdot DE$