Giải thích các bước giải:

a.Xét $\Delta ABC, \Delta HAC$ có:

Chung $\hat C$

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

$\to \Delta ABC\sim\Delta HAC(g.g)$ 

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

$\to AC^2=HC.CB$

b.Tương tự $a\to CA^2=CE.CD$

$\to CE.CD=CH.CB$

$\to \dfrac{CE}{CH}=\dfrac{CB}{CD}$

$\to \Delta CHE\sim\Delta CDB(c.g.c)$

$\to \widehat{CEH}=\widehat{CBD}=\widehat{ABC}$

$\to \widehat{KED}=\widehat{CEH}=\widehat{ABC}=\widehat{KBH}$

$\to \Delta KED\sim\Delta KBH(g.g)$

$\to \dfrac{KE}{KB}=\dfrac{KD}{KH}$

$\to KD.KB=KE.KH$

Ta có:

$\widehat{KAE}=\widehat{DAE}=90^o-\widehat{EAC}=\widehat{ECA}$

Gọi $EC\cap AH=F$

$\to \widehat{FEA}=\widehat{FHC}(=90^o)$

$\to \Delta FEA\sim\Delta FHC(g.g)$

$\to \dfrac{FE}{FH}=\dfrac{FA}{FC}$

$\to \Delta FEH\sim\Delta FAC(c.gc.)$

$\to \widehat{FCA}=\widehat{FHE}$

$\to \widehat{KAE}=\widehat{KHA}$

$\to \Delta KAE\sim\Delta KHA(g.g)$

$\to \dfrac{KA}{KH}=\dfrac{KE}{KA}$

$\to KA^2=KH.KE$

$\to AK^2=KD.KB$