Giải thích các bước giải:

1.Xét $\Delta HEC,\Delta HBF$ có:

$\widehat{EHC}=\widehat{FHB}$

$\widehat{HEC}=\widehat{HFB}(=90^o)$

$\to\Delta HEC\sim\Delta HFB(g.g)$

$\to\dfrac{HE}{HF}=\dfrac{HC}{HB}$

$\to HE.HB=HF.HC$

$\to\dfrac{HE}{HC}=\dfrac{HF}{HB}$

Mà $\widehat{FHE}=\widehat{BHC}$

$\to\Delta HEF\sim\Delta HCB(c.g.c)$

$\to\widehat{HEF}=\widehat{HCB}$

$\to\widehat{BEF}=\widehat{BCF}$

2.Ta có $BE\perp AC, CF\perp AB, BE\cap CH=H$

$\to H$ là trực tâm $\Delta ABC\to AH\perp BC\to AK\perp BC$

Xét $\Delta BKA,\Delta BFC$ có:

Chung $\hat B$

$\widehat{BKA}=\widehat{BFC}(=90^o)$

$\to\Delta BKA\sim\Delta BFC(g.g)$

$\to\dfrac{BK}{BF}=\dfrac{BA}{BC}$

$\to BK.BC=BF.BA$

Tương tự $CE.CA=CK.CB, AE.AC=AF.AB$

Ta có:

$2(AE.AC+CK.CB+BF.BA)$

$=(AE.AC+CK.CB)+(AE.AC+BF.BA)+(CK.CB+BF.BA)$

$=(AE.AC+CE.CA)+(AF.AB+BF.BA)+(CK.CB+BK.BC)$

$=AC^2+AB^2+BC^2$

$\to đpcm$

3.Xét $\Delta BHK,\Delta BEC$ có:

Chung $\hat B$

$\widehat{BKH}=\widehat{BEC}(=90^o)$

$\to\Delta BHK\sim\Delta BCE(g.g)$

$\to\dfrac{BH}{BC}=\dfrac{BK}{BE}$

$\to\dfrac{BH}{BK}=\dfrac{BC}{BE}$

Mà $\widehat{EBK}=\widehat{HBC}$

$\to\Delta BKE\sim\Delta BHC(c.g.c)$

$\to\widehat{BEK}=\widehat{BCH}=\widehat{BCF}=\widehat{BEF}$

$\to EH$ là phân giác $\widehat{JEK}$

Mà $EA\perp EH$

$\to EA$ là phân giác ngoài tại đỉnh $E$ của $\Delta EJK$

$\to \dfrac{HJ}{HK}=\dfrac{AJ}{AK}$

$\to\dfrac{HJ}{AJ}=\dfrac{HK}{AK}$

3b.Ta cần chứng minh:

$\dfrac{2}{KJ}=\dfrac1{KH}+\dfrac1{AK}$

$\leftrightarrow \dfrac{2}{KJ}=\dfrac1{KH}+\dfrac1{AK}$

$\leftrightarrow \dfrac{1}{KJ}-\dfrac1{AK}=\dfrac1{KH}-\dfrac{1}{KJ}$

$\leftrightarrow \dfrac{AK-KJ}{KJ\cdot AK}=\dfrac{KJ-KH}{KJ\cdot KH}$

$\leftrightarrow \dfrac{AJ}{KJ\cdot AK}=\dfrac{HJ}{KJ\cdot KH}$

$\leftrightarrow \dfrac{AJ}{ AK}=\dfrac{HJ}{KH}$ luôn đúng vì $\dfrac{HJ}{AJ}=\dfrac{HK}{AK}$

$\to đpcm$

4.Ta có:

$\dfrac{HK}{AK}=\dfrac{S_{HBC}}{S_{ABC}}$

$\dfrac{HE}{BE}=\dfrac{S_{HAC}}{S_{ABC}}$

$\dfrac{HF}{CF}=\dfrac{S_{AHB}}{S_{ABC}}$

$\to \dfrac{HK}{AK}+\dfrac{HE}{BE}+\dfrac{HF}{CF}=\dfrac{S_{HBC}+S_{HAC}+S_{AHB}}{S_{ABC}}=\dfrac{S_{ABC}}{S_{ABC}}=1$

5.Xét $\Delta KHB,\Delta KAC$ có:

$\widehat{HKB}=\widehat{AKC}(=90^o)$

$\widehat{BHK}=90^o-\widehat{HBK}=90^o-\widehat{EBC}=\widehat{ECB}=\widehat{ACK}$

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

$\to\dfrac{KB}{KA}=\dfrac{KH}{KC}$

$\to  KH.KA=KB.KC\le\dfrac14(KB+KC)^2=\dfrac14BC^2=\dfrac14a^2$

Dấu = xảy ra khi $KB=KC\to K$ là trung điểm $BC\to AK\perp BC$ tại trung điểm $BC$

$\to\Delta ABC$ cân tại $A$