Giải thích các bước giải:
a.Xét $\Delta CHB,\Delta CKD$ có:
$\widehat{CHB}=\widehat{CKD}(=90^o)$
$\widehat{CBH}=\widehat{BAD}=\widehat{KDC}$
$\to\Delta CHB\sim\Delta CKD(g.g)$
$\to\dfrac{CH}{CK}=\dfrac{CB}{CD}$
$\to \dfrac{CH}{CB}=\dfrac{CK}{CD}$
b. Ta có $CH\perp AB, CK\perp AD$
$\to \widehat{HCK}=360^o-\widehat{AHC}-\widehat{AKC}-\widehat{BAD}=180^o-\widehat{BAD}=\widehat{ABC}$
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 $BE\perp AC, DF\perp AC$
Do $ABCD$ là hình bình hành
$\to S_{ABC}=S_{ACD}$
$\to\dfrac12BE\cdot AC=\dfrac12DF\cdot AC$
$\to BE=DF$
$\to AF=\sqrt{AD^2-DF^2}=\sqrt{BC^2-BE^2}=CE$
Xét $\Delta ABE,\Delta ACH$ có:
Chung $\hat A$
$\widehat{AEB}=\widehat{AHC}(=90^o)$
$\to\Delta ABE\sim\Delta ACH(g.g)$
$\to\dfrac{AB}{AC}=\dfrac{AE}{AH}$
$\to AB.AH=AE.AC$
Tương tự $AD.AK=AF.AC$
$\to AB.AH+AD.AK=AE.AC+AF.AC=(AE+AF).AC=(AE+CE).AC=AC62$
$\to đpcm$
d.Ta có: $AB//BC, AB//CD$
$\to \dfrac{ID}{IM}=\dfrac{IC}{IA}=\dfrac{IN}{ID}$
$\to ID^2=IM.IN$
e.Ta có $D, J$ đối xứng qua $I\to ID=IJ\to IJ^2=IM.IN$
Ta có:
$\dfrac{DI}{DM}=\dfrac{CI}{CA}$
$\dfrac{DI}{DN}=\dfrac{AI}{AC}$
$\to \dfrac{DI}{DN}:\dfrac{DI}{DM}=\dfrac{AI}{AC}:\dfrac{CI}{CA}$
$\to \dfrac{DM}{DN}=\dfrac{AI}{CI}$
Ta có:
$\dfrac{JM}{JN}=\dfrac{IM-IJ}{IJ-IN}$
$\to\dfrac{JM}{JN}=\dfrac{IM-ID}{ID-IN}$
Mà $ID^2=IM.IN$
$\to \dfrac{ID}{IM}=\dfrac{IN}{ID}=\dfrac{ID-IN}{IM-ID}=\dfrac{JN}{JM}$
Mà $\dfrac{ID}{IM}=\dfrac{CI}{AI}$
$\to \dfrac{CI}{AI}=\dfrac{JN}{JM}$
$\to \dfrac{JM}{JN}=\dfrac{AI}{CI}$
$\to \dfrac{JM}{JN}=\dfrac{DM}{DN}$
$\to đpcm$
