Giải thích các bước giải:
a.Xét $\Delta ABC, \Delta HBA$ có:
Chung $\hat B$
$\widehat{BAC}=\widehat{AHB}(=90^o)$
$\to \Delta ABC\sim\Delta HBA(g.g)$
b.Xét $\Delta AHB,\Delta AHC$ có:
$\widehat{AHB}=\widehat{AHC}(=90^o)$
$\widehat{BAH}=90^o-\widehat{HAC}=\widehat{ACH}$
$\to\Delta AHB\sim\Delta CHA(g.g)$
$\to \dfrac{AH}{CH}=\dfrac{HB}{HA}$
$\to AH^2=HB.HC$
c.Ta có $\Delta ABC$ vuông tại $A\to BC^2=AB^2+AC^2=225$
$\to BC=15$
Vì $AH\perp BC\to AH\cdot BC=AB\cdot AC(=2S_{ABC})$
$\to AH=\dfrac{AB\cdot AC}{BC}=\dfrac{36}{5}$
$\to CH=\sqrt{AC^2-AH^2}=\dfrac{48}{5}$
Xét $\Delta ADC, \Delta HCE$ có:
$\widehat{DAC}=\widehat{EHC}(=90^o)$
$\widehat{ACD}=\widehat{ECH}$
$\to\Delta CAD\sim\Delta CHE(g.g)$
$\to \dfrac{S_{ACD}}{S_{HCE}}=(\dfrac{CA}{CH})^2=\dfrac{25}{16}$
