Giải thích các bước giải:

a.Xét $\Delta CDE, \Delta CAB$ có:

Chung $\hat C$

$\widehat{CDE}=\widehat{CAB}(=90^o)$

$\to\Delta CDE\sim\Delta CAB(g.g)$

$\to \dfrac{DE}{AB}=\dfrac{CD}{CA}$

$\to DE=\dfrac{AB\cdot CD}{AC}$

Ta có $AD$ là phân giác $\hat A$

$\to \dfrac{DC}{DB}=\dfrac{AC}{AB}=\dfrac43$

$\to \dfrac{DC}{DC+DB}=\dfrac4{3+4}=\dfrac47$

$\to \dfrac{DC}{BC}=\dfrac47$

$\to CD=\dfrac47BC$

$\to CD=\dfrac{100}7$

$\to DE=\dfrac{15\cdot \dfrac{100}7}{20}=\dfrac{75}7$

b.Từ câu a 

$\to \dfrac{CE}{BC}=\dfrac{CD}{BA}$

$\to AB.CE=CD.BC$

c.Ta có:

$S_{CDE}=\dfrac12DE\cdot CD=\dfrac12\cdot \dfrac{75}7\cdot \dfrac{100}7=\dfrac{3750}{49}$

$\to S_{AEDB}=S_{ABC}-S_{CDE}=\dfrac12AB\cdot AC-\dfrac{3750}{49}$

$\to S_{AEDB}=\dfrac12\cdot 15\cdot 20-\dfrac{3750}{49}$

$\to S_{AEDB}=\dfrac{3600}{49}$