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:
$S_{CDE}=\dfrac12DE\cdot CD=\dfrac12\cdot \dfrac{75}7\cdot \dfrac{100}7=\dfrac{3750}{49}$
$\to S_{AEDB}=S_{ABC}-S_{CDE}=\dfrac12AB\cdot AC-\dfrac{3750}{49}$
$\to S_{AEDB}=\dfrac12\cdot 15\cdot 20-\dfrac{3750}{49}$
$\to S_{AEDB}=\dfrac{3600}{49}$