Giải thích các bước giải:
a.Xét $\Delta ABC, \Delta HBA$ có:
Chung $\hat B$
$\widehat{BAC}=\widehat{BHA}(90^o)$
$\to \Delta ABC\sim\Delta HBA(g.g)$
b.Ta có $\Delta ABC$ vuông tại $A\to BC=\sqrt{AB^2+AC^2}=25$
Vì $BD$ là phân giác $\hat B$
$\to \dfrac{DA}{DC}=\dfrac{BA}{BC}=\dfrac35$
$\to\dfrac{DA}{DA+DC}=\dfrac3{3+5}$
$\to \dfrac{DA}{AC}=\dfrac38$
$\to AD=\dfrac38AC=\dfrac{15}{2}$
$\to CD=AC-AD=\dfrac{25}{2}$
c.Xét $\Delta CKD, \Delta ABC$ có:
Chung $\hat C$
$\widehat{CKD}=\widehat{CAB}(=90^o)$
$\to \Delta CDK\sim\Delta CBA(g.g)$
$\to \dfrac{S_{CDK}}{S_{CBA}}=(\dfrac{CD}{CB})^2=\dfrac14$
$\to S_{ABC}=4S_{KDC}$
d.Xét $\Delta ADB, \Delta BEH$ có:
$\widehat{BAD}=\widehat{BHE}$
$\widehat{ABD}=\widehat{EBH}$
$\to \Delta BAD\sim\Delta BHE(g.g)$
$\to \widehat{ADB}=\widehat{BEH}$
$\to \widehat{AED}=\widehat{BEH}=\widehat{ADB}=\widehat{ADE}$
$\to \Delta ADE$ cân tại $A\to AE=AD=\dfrac{15}2$
Ta có $\Delta ABC$ vuông tại $A, AH\perp BC$
$\to S_{ABC}=\dfrac12AH\cdot BC=\dfrac12AB\cdot AC$
$\to AH=\dfrac{AB\cdot AC}{BC}=12$
$\to \dfrac{AE}{AH}=\dfrac58$
