Giải thích các bước giải:

a.Xét $\Delta AHB,\Delta AHC$ có:

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

$\widehat{HAB}=90^o-\widehat{HAC}=\hat C$

$\to \Delta HAB\sim\Delta HCA(g.g)$

$\to \dfrac{HA}{HC}=\dfrac{HB}{HA}$

$\to HA^2=HB.HC$

$\to HA.HK=HB.HC$

b.Ta có: $AK\perp BC=H$ là trung điểm $AK$

$\to BC$ là trung trực $AK$

$\to BA=BK$

$\to \Delta ABK$ cân tại $B$

Ta có: $AM//BC$

$\to \widehat{TAM}=\widehat{TCB}=90^o-\widehat{ABC}=90^o-\widehat{ABH}=\widehat{HAB}=\widehat{KAB}=\widehat{AKB}$

$\to \Delta TAM\sim\Delta TKA(g.g)$

$\to \dfrac{TA}{TK}=\dfrac{TM}{TA}$

$\to TA^2=TM.TK$

Ta có: $AM//BC, AH\perp BC\to AM\perp AH$

$\to \widehat{MAN}=90^o-\widehat{NAK}=\widehat{AKN}=\widehat{AKT}=\widehat{TAM}$

$\to AM$ là phân giác $\widehat{TAN}$

Do $AM\perp AK$

$\to AK$ là phân giác ngoài tại $A$ của $\Delta ATN$

$\to \dfrac{MT}{MN}=\dfrac{KT}{KN}$

$\to MT.KN=MN.KT$

c.Gọi $KC\cap AM=D$

Vì $BC$ là trung trực $AK$

$\to BA=BK, CA=CK$

$\to \Delta ABC=\Delta KBC(c.c.c)$

$\to \widehat{BKC}=\widehat{BAC}=90^o$

$\to BK\perp KC$

$\to AN//KC$

$\to AN//KD$

Ta có: $CA=CK\to \Delta CKA$ cân tại $C$

$\to \widehat{CAD}=90^o-\widehat{CAK}=90^o-\widehat{CKA}=\hat D$

$\to \Delta CAD$ cân tại $C$

$\to CA=CD$

$\to CK=CD$

Vì $AN//DK$

$\to \dfrac{EN}{CK}=\dfrac{ME}{MC}=\dfrac{AE}{CD}$

$\to EN=EA$

d.Gọi $AK\cap MQ=G$

$\to \widehat{GQK}=\widehat{GAM}(=90^o)$

     $\widehat{QGK}=\widehat{AGM}$

$\to \Delta GQK\sim\Delta GAM(g.g)$

$\to \dfrac{GQ}{GA}=\dfrac{GK}{GM}$

Mà $\widehat{AGQ}=\widehat{MGK}$

$\to \Delta GAQ\sim\Delta GMK(g.g)$

$\to \widehat{AQG}=\widehat{GKM}$

$\to \widehat{FQC}=\widehat{AQG}=\widehat{GKM}=\widehat{TKA}=\widehat{ACB}=\widehat{FCA}$

$\to \Delta FQC\sim\Delta FCA(g.g)$

$\to \dfrac{FQ}{FC}=\dfrac{FC}{FA}$

$\to FC^2=FQ.FA$

Gọi $KQ\cap CH=I$

Ta có: $KQ\perp MC$

$\to \widehat{IQC}=\widehat{IHK}(=90^o)$

      $\widehat{QIC}=\widehat{HIK}$

$\to \Delta IQC\sim\Delta IHK(g.g)$

$\to \dfrac{IQ}{IH}=\dfrac{IC}{IK}$

Mà $\widehat{HIQ}=\widehat{KIC}$

$\to \Delta IQH\sim\Delta ICK(c.g.c)$

$\to \widehat{HQI}=\widehat{ICK}$

$\to \widehat{HQK}=\widehat{HCK}=\widehat{ACH}$

Ta có: 

$\widehat{IQF}=90^o-\widehat{AQG}=90^o-\widehat{AKB}=90^o-\widehat{ACH}=90^o-\widehat{HQK}$

$\to \widehat{IQF}+\widehat{HQK}=90^o$

$\to \widehat{HQF}=90^o$

$\to \widehat{HQF}=\widehat{FHA}$

$\to \Delta FQH\sim\Delta FHA(g.g)$

$\to \dfrac{FQ}{FH}=\dfrac{FH}{FA}$

$\to FH^2=FA.FQ$

$\to FH^2=FC^2$

$\to FH=FC$