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 $\widehat{CED}=\widehat{CAB}=90^o$

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

$\to \dfrac{S_{CDE}}{S_{CBA}}=(\dfrac{CD}{CB})^2$

$\to \dfrac{S_{CDE}}{S_{CBA}}=(\dfrac{CB-BD}{CB})^2$

$\to \dfrac{S_{CDE}}{S_{CBA}}=(\dfrac{CB-2BH}{CB})^2$

$\to \dfrac{S_{CDE}}{S_{CBA}}=(\dfrac{CB-2\cdot\dfrac{AB^2}{BC}}{CB})^2$

$\to S_{CDE}=(\dfrac{CB-2\cdot\dfrac{AB^2}{BC}}{CB})^2\cdot S_{ABC}$