Giải thích các bước giải:
a.Vì $CF,AD$ là đường cao $\Delta ABC$
$\to CF\perp AB,AD\perp BC$
$\to \widehat{AFH}=\widehat{ADB}=90^o$
$\to \Delta AFH\sim\Delta ADB(g.g)$
b.Do $CF\perp AB,BE\perp AC\to \widehat{CEH}=\widehat{CFA}=90^o$
$\to \Delta CEH\sim\Delta CFA(g.g)$
$\to \dfrac{CE}{CF}=\dfrac{CH}{CA}\to CE.CA=CF.CH$
c.Ta có $\widehat{AEH}=\widehat{HDB}=90^o,\widehat{AHE}=\widehat{BHD}$
$\to \Delta AEH\sim\Delta BDH(g.g)$
$\to \dfrac{AH}{BH}=\dfrac{EH}{DH}$
$\to AH.DH= BH.HE$
d.Ta có :
$\widehat{IAH}=\widehat{FAH}=90^o-\widehat{AHF}=90^o-\widehat{DHC}=\widehat{DCH}=\widehat{HCM}$
$\widehat{AIH}=\widehat{FIH}=90^o-\widehat{FHI}=90^o-\widehat{KHC}=\widehat{MHC}$ vì $HM\perp IK$
$\to \Delta AHI\sim\Delta CMH(g.g)$
$\to \dfrac{AH}{CM}=\dfrac{HI}{MH}$
$\to HI=\dfrac{AH.MH}{CM}$
Tương tự $\to HK=\dfrac{AH.MH}{BM}$
Mà $M $ là trung điểm BC$\to MB=MC$
$\to \dfrac{AH.MH}{CM}=\dfrac{AH.MH}{BM}\to HI=HK$
