@1926877
b/ \(ΔAHC\backsim ΔBAC→\dfrac{HC}{AC}=\dfrac{AC}{BC}↔AC^2=HC.BC\)
Xét \(ΔCKA\) và \(ΔCAD\):
\(\widehat C:chung\)
\(\widehat{CKA}=\widehat{CAD}(=90^\circ)\)
\(→ΔCKA\backsim ΔCAD(g-g)\)
\(→\dfrac{CK}{CA}=\dfrac{CA}{CD}\)
\(↔CA^2=CK.CD\) mà \(AC^2=HC.BC\)
\(→CK.CD=HC.BC↔\dfrac{CK}{CH}=\dfrac{CB}{CD}\)
Xét \(ΔCHK\) và \(ΔCDB\):
\(\widehat C:chung\)
\(\dfrac{CK}{CH}=\dfrac{CB}{CD}(cmt)\)
\(→ΔCHK\backsim ΔCDB(c-g-c)\)
c/ Xét \(ΔAHB\) và \(ΔCAB\):
\(\widehat{AHB}=\widehat{CAB}(=90^\circ)\)
\(\widehat B:chung\)
\(→ΔAHB\backsim ΔCAB(g-g)\)
\(→\dfrac{AH}{AB}=\dfrac{AC}{BC}↔AB.AC=AH.BC\)
Xét \(ΔAKD\) và \(ΔCAD\):
\(\widehat D:chung\)
\(\widehat{AKD}=\widehat{CAD}(=90^\circ)\)
\(→ΔAKD\backsim ΔCAD(g-g)\)
\(→\dfrac{AK}{AD}=\dfrac{CA}{DC}↔AK.DC=AC.AD\)
Xét \(ΔBAC\) và \(ΔBCD\):
\(\widehat B:chung\)
\(\widehat{BAC}=\widehat{BCD}(=90^\circ)\)
\(→ΔBAC\backsim ΔBCD(g-g)\)
\(→\dfrac{CA}{CB}=\dfrac{DC}{DB}↔CA.BD=CB.CD\)
Xét tứ giác \(AHCK\): \(\widehat{H}=\widehat{C}=\widehat{K}=90^\circ\)
\(→AHCK\) là hình chữ nhật
\(→CK=AH\) và \(CH=AK\)
\(\dfrac{CK}{CD}+\dfrac{CH}{CB}\\=\dfrac{CK.CB+CD.CH}{CD.CB}\\=\dfrac{AH.CB+AK.CD}{CA.BD}\\=\dfrac{AB.AC+AC.AD}{CA.BD}\\=\dfrac{AC(AB+AD)}{CA.BD}\\=\dfrac{AC.BD}{CA.BD}\\=1\)
