Giải thích các bước giải:

a.Ta có $\Delta ABC$ vuông tại $A$

$\to BC^2=AB^2+AC^2$

$\to AC^2=BC^2-AB^2=400$

$\to AC=20$

b.Xét $\Delta HAC,\Delta ABC$  có:

Chung $\hat C$

$\widehat{AHC}=\widehat{BAC}(=90^o)$

$\to \Delta CHA\sim\Delta CAB(g.g)$

$\to \dfrac{CH}{CA}=\dfrac{CA}{BC}$

$\to CH=\dfrac{AC^2}{BC}=16$

$\to HB=BC-HB=9$

Mặt khác $\dfrac{HA}{AB}=\dfrac{AC}{BC}$

$\to AH=\dfrac{AC\cdot AB}{BC}=12$

c.Từ câu b$\to \widehat{HAC}=\widehat{ABC}=\widehat{ABH}$

Xét $\Delta AHB, \Delta AHC$ có:

$\widehat{AHB}=\widehat{AHC}(=90^o)$

$\widehat{HAC}=\widehat{ABH}$

$\to\Delta AHB\sim\Delta CHA(g.g)$
$\to \dfrac{AH}{CH}=\dfrac{HB}{HA}$

$\to AH^2=HB.HC$

d, e Thiếu dữ kiện

f.Xét $\Delta AHB ,\Delta ABC$ có:

Chung $\hat B$

$\widehat{AHB}=\widehat{BAC}=90^o$

$\to \Delta BAH\sim\Delta BCA(g.g)$

$\to \dfrac{AH}{AC}=\dfrac{BH}{BA}$

$\to \dfrac{2AE}{AC}=\dfrac{BH}{\dfrac12BD}$

$\to \dfrac{AE}{AC}=\dfrac{BH}{BD}$

$\to \dfrac{AE}{BH}=\dfrac{AC}{BD}$

Mà $\widehat{EAC}=\widehat{HAC}=\widehat{ABC}=\widehat{DBH}$

$\to \Delta HBD\sim\Delta EAC(c.g.c)$

g.Từ câu g

$\to \widehat{ACE}={BDH}\to \widehat{ICK}=\widehat{IDA}$

Mà $\widehat{DIA}=\widehat{KIC}$

$\to\Delta ADI\sim\Delta KIC(g.g)$

$\to \widehat{IKC}=\widehat{IAD}=90^o$

$\to EC\perp DH$

Kẻ $IF\perp CD$

Xét $\Delta CFI, \Delta CAD$ có:

Chung $\hat C$

$\widehat{CFI}=\widehat{CAD}(=90^o)$

$\to\Delta CIF\sim\Delta CDA(g.g)$

$\to \dfrac{CI}{CD}=\dfrac{CF}{CA}$

$\to CI.CA=CF.CD$

Tương tự chứng minh được $DI.DK=DF.DC$

$\to CI.CA+DI.DK=CF.CD+DF.DC=DC^2$