Giải thích các bước giải:
a.Ta có : $CF\perp AB, BE\perp AC\to\widehat{HEA}+\widehat{HFA}=90^o+90^o=180^o$
$\to AEHF$ nội tiếp
b.Ta có : $\widehat{FHA}=\widehat{DHC},\widehat{AFH}=\widehat{HDC}=90^o\to\widehat{BAH}=\widehat{FCB}$
$\to\Delta HDC\sim\Delta BDA(g.g)$
$\to\dfrac{HD}{BD}=\dfrac{DC}{DA}\to DH.DA=DB.DC$
c.Ta có :
$\widehat{OAC}=90^o-\dfrac 12\widehat{AOC}=90^o-\widehat{ABC}=\widehat{FAH}=\widehat{HEF}$
$\to AO\perp EF$
Vì $HD\perp BC, DI\perp BH, DK\perp CH$
$\to HI.HB=HD^2=HK.HC\to \dfrac{HI}{HC}=\dfrac{HK}{HB}\to\Delta HIK\sim\Delta HCB(c.g.c)$
$\to\widehat{HIK}=\widehat{HCB}$
Mà $\widehat{BFC}=\widehat{BEC}=90^o\to BFEC$ nội tiếp
$\to \widehat{HIK}=\widehat{HCB}=\widehat{FEB}\to EF//IK\to IK\perp AO$
d.
