Giải thích các bước giải:

1.Ta có: $\Delta ABC$ vuông tại $A,AJ\perp BC$

$\to\widehat{AJB} =\widehat{AJC}=90^o,\widehat{BAJ}=90^o-\widehat{JAC}=\widehat{ACJ}$

$\to\Delta AJB\sim\Delta CJA(g.g)$

$\to\dfrac{AJ}{CJ}=\dfrac{JB}{JA}$

$\to AJ^2=JB.JC$

Ta có: $DH\perp AB, HK\perp AC, AB\perp AC$

$\to AHDK$ là hình chữ nhật

Mà $AD$ là phân giác góc $A\to AHDK$ là hình vuông

$\to DH//AC, DK//AB, AH=HD=DK=AK$

$\to \dfrac{HM}{AK}=\dfrac{BM}{BK}=\dfrac{MD}{CK}$

$\to \dfrac{HM}{MD}=\dfrac{KA}{KC}=\dfrac{DK}{KC}$

Lại có: $\widehat{BHD}=\widehat{DKC}=90^o,\widehat{HDB}=\widehat{KCD}$

$\to \Delta HBD\sim\Delta KDC(g.g)$

$\to \dfrac{HB}{DK}=\dfrac{HD}{KC}$

$\to \dfrac{DK}{CK}=\dfrac{HB}{HD}$

$\to \dfrac{HM}{MD}=\dfrac{HB}{HD}=\dfrac{HB}{HA}$

2.Ta có: $\dfrac{HM}{MD}=\dfrac{KA}{KC}=\dfrac{NH}{NC}$

$\to MN//CD\to MN//BC$

3.Ta có: $\widehat{AKN}=\widehat{BAK}=90^o$

$\dfrac{KN}{KA}=\dfrac{KN}{AH}=\dfrac{CK}{CA}=\dfrac{KD}{AB}=\dfrac{KA}{AB}$

$\to \Delta KAN\sim\Delta ABK(c.g.c)$

$\to \widehat{KAN}=\widehat{ABK}$

Gọi $AN\cap BK=E\to \widehat{KAE}=\widehat{KAB}$

Mà $\widehat{EKA}=\widehat{BKA}$

$\to\Delta KAE\sim\Delta KBA(g.g)$

$\to \widehat{KEA}=\widehat{KAB}=90^o$

$\to AE\perp BK\to MI\perp AN$

Chứng minh tương tự $NI\perp AM$

$\to I$ là trực tâm $\Delta AMN\to AI\perp MN$

Mà $MN//BC\to AI\perp BC$

$\to A,I,J$ thẳng hàng