Giải thích các bước giải:
a.ta có $AH\perp BC\to\widehat{AHB}=\widehat{BAC}=90^o$
$\to\Delta ABC\sim\Delta HBA(g.g)$
b.Từ câu a $\to\dfrac{AB}{BH}=\dfrac{BC}{AB}\to BH=\dfrac{AB^2}{BC}$
Ta có $BC=\sqrt{AB^2+AC^2}=50$
$\to BH=18$
c.Vì $HB=HD, AH\perp BC=H\to AH$ là trung trực của $BH$
Mà $EC\perp AD$
$\to\widehat{DCE}=90^o-\widehat{EDC}=90^o-\widehat{ADH}=\widehat{DAH}=\widehat{BAH}=\widehat{ACB}=\widehat{ACH}$
$\to CD$ là phân giác $\widehat{ACE}$
$\to\dfrac{DA}{DE}=\dfrac{CA}{CE}$
$\to\dfrac{AB}{DE}=\dfrac{CA}{CE}$
$\to AB.EC=AC.ED$
d.Từ câu c
$\to\widehat{DCE}=\widehat{ACB}$
Mà $\widehat{CED}=\widehat{CAB}=90^o$
$\to\Delta CDE\sim\Delta CBA(g.g)$
$\to \dfrac{S_{CDE}}{S_{CBA}}=(\dfrac{CD}{CB})^2$
$\to \dfrac{S_{CDE}}{S_{CBA}}=(\dfrac{CB-BD}{CB})^2$
$\to \dfrac{S_{CDE}}{S_{CBA}}=(\dfrac{CB-2BH}{CB})^2$
$\to \dfrac{S_{CDE}}{S_{CBA}}=(\dfrac{CB-2\cdot\dfrac{AB^2}{BC}}{CB})^2$
$\to S_{CDE}=(\dfrac{CB-2\cdot\dfrac{AB^2}{BC}}{CB})^2\cdot S_{ABC}$
