Giải thích các bước giải:

a.Ta có :

$\widehat{CKD}=\widehat{CHB}=90^o,\widehat{KDC}=180^o-\widehat{ADC}=180^o-\widehat{ABC} =\widehat{HBC}$

$\to \Delta CDK\sim\Delta CBH(g.g)$

$\to \dfrac{CD}{CB}=\dfrac{CK}{CH}\to\dfrac{CH}{CB}=\dfrac{CK}{CD}$

b.Ta có:
$\widehat{HCK}=\widehat{HCA}+\widehat{KCA}=90^o-\widehat{CAH}+90^o-\widehat{CAK}=180^o-\widehat{HAK}=180^o-\widehat{DAB}=\widehat{ABC}$

Mà $\dfrac{CH}{CB}=\dfrac{CK}{CD}\to \dfrac{CH}{CB}=\dfrac{CK}{AB}$

$\to\Delta CHK\sim\Delta BCA(c.g.c)$

c.Kẻ $BF\perp AC, DE\perp AC$

$\to \widehat{AED}=\widehat{BFC}=90^o$
Mà $AD=CB,\widehat{DAE}=\widehat{DAC}=\widehat{ACB}=\widehat{FCB}$

$\to\Delta AED=\Delta CFB$(cạnh huyền-góc nhọn)

$\to AE=CF$

Ta có $\widehat{AFB}=\widehat{AHC}=90^o,\widehat{BAF}=\widehat{HAC}$

$\to\Delta AFB\sim\Delta AHC(g.g)$

$\to\dfrac{AF}{AH}=\dfrac{AB}{AC}$

$\to AB.AH=AF.AC$

Tương tự $AD.AK=AE.AC=CF.AC$

$\to AB.AH+AD.AK=AF.AC+CF.AC=AC^2$

d.Ta có :

$AM//CD\to\dfrac{IM}{ID}=\dfrac{IA}{IC}$
Mà $AD//CN\to\dfrac{IA}{IC}=\dfrac{ID}{IN}$

$\to\dfrac{IM}{ID}=\dfrac{ID}{IN}$

$\to ID^2=IM.IN$

e.Ta có $J,D$ đối xứng qua $I$

$\t ID=IJ$

$\to IJ^2=IM.IN$

$\to IJ.IJ=IM.IN$

$\to (MI+MJ).(NI-NJ)=MI.NI$

$\to MI.NI+MJ.NI-MI.NJ-MJ.NJ=MI.NI$

$\to MJ.NI=MI.NJ+MJ.NJ$

$\to\dfrac{MJ}{NJ}\cdot NI=MI+MJ$

$\to\dfrac{MJ}{NJ}\cdot NI=IJ$

$\to\dfrac{MJ}{NJ}=\dfrac{IJ}{NI}=\dfrac{ID}{IN}$

Lại có:

$AM//CD\to \dfrac{ID}{IM}=\dfrac{IC}{IA}\to\dfrac{ID}{ID+IM}=\dfrac{IC}{IA+IC}\to\dfrac{ID}{DM}=\dfrac{IC}{AC}$

$AD//CN\to\dfrac{ID}{IN}=\dfrac{IA}{IC}\to\dfrac{ID+IN}{IN}=\dfrac{IA+IC}{IC}\to\dfrac{DN}{IN}=\dfrac{AC}{IC}$

$\to \dfrac{ID}{DM}\cdot\dfrac{DN}{IN}=\dfrac{IC}{AC}\cdot\dfrac{AC}{IC}=1$

$\to \dfrac{DM}{DN}=\dfrac{ID}{IN}$

$\to\dfrac{MJ}{NJ}=\dfrac{DM}{DN}$