Giải thích các bước giải:
a.Xét $\Delta AHB, \Delta ABC$ có:
Chung $\hat B$
$\widehat{AHB}=\widehat{BAC}(=90^o)$
$\to \Delta ABH\sim\Delta CBA(g.g)$
$\to \dfrac{AB}{CB}=\dfrac{BH}{BA}$
$\to AB^2=BH.BC$
b.Ta có $AC=\sqrt{BC^2-AB^2}=20, AF=AB+BF=20$
Từ câu a $\to BH=\dfrac{AB^2}{BC}=9$
Ta có $EF\perp AH\to EF//HB$
$\to \dfrac{BH}{EF}=\dfrac{AB}{AF}=\dfrac34$
$\to EF=\dfrac43BH$
$\to EF=12$
c.Gọi $AB\cap EK=D$
Xét $\Delta DEB,\Delta DAK$ có:
Chung $\hat D$
$\widehat{DEB}=\widehat{DAK}(=90^o)$
$\to \Delta DEB\sim\Delta DAK(g.g)$
$\to \dfrac{DE}{DA}=\dfrac{DB}{DK}$
$\to \dfrac{DE}{DB}=\dfrac{DA}{DK}$
Mà $\widehat{ADE}=\widehat{BDK}$
$\to \Delta DEA\sim\Delta DBK(c.g.c)$
$\to \widehat{DAE}=\widehat{DKB}$
$\to \widehat{FAE}=\widehat{BKE}$
Mà $\widehat{AEF}=\widehat{KEB}(=90^o)$
$\to \Delta AEF\sim\Delta KEB(g.g)$
$\to \dfrac{AE}{EK}=\dfrac{EF}{EB}=\dfrac{AF}{BK}$
$\to AF.BE=BK.EF$