a/ Xét \(ΔHCA\) và \(ΔACB\):
\(\widehat{AHC}=\widehat{BAC}(=90^\circ)\)
\(\widehat C:chung\)
\(→ΔHCA\backsim ΔACB(g-g)\)
\(→\dfrac{CH}{CA}=\dfrac{CA}{CB}\)
\(↔CA^2=CH.CB\)
b/ Xét \(ΔAHB\) và \(ΔCHA\):
\(\widehat{AHB}=\widehat{CHA}(=90^\circ)\)
\(\widehat{ABH}=\widehat{CAH}\) (cùng phụ \(\widehat C\) )
\(→ΔAHB\backsim ΔCHA(g-g)\)
\(→\dfrac{AH}{HB}=\dfrac{CH}{AH}\)
\(↔AH^2=HB.CH\)
c/ Xét \(ΔAHF\) và \(ΔACH\):
\(\widehat{AFH}=\widehat{AHC}(=90^\circ)\)
\(\widehat{HAF}:chung\)
\(→ΔAHF\backsim ΔACH(g-g)\)
\(→\dfrac{AH}{AF}=\dfrac{AC}{AH}\)
\(↔AH^2=AF.AC\) (1)
Xét \(ΔAHE\) và \(ΔABH\):
\(\widehat{AEH}=\widehat{AHB}(=90^\circ)\)
\(\widehat{HAE}:chung\)
\(→ΔAHE\backsim ΔABH(g-g)\)
\(→\dfrac{AH}{AE}=\dfrac{AB}{AH}\)
\(↔AH^2=AE.AB\) (2)
(1)(2) \(→AE.AB=AF.AC\)
\(→\dfrac{AB}{AF}=\dfrac{AC}{AE}\)
Xét \(ΔABF\) và \(ΔACE\):
\(\dfrac{AB}{AF}=\dfrac{AC}{AE}(cmt)\)
\(\widehat A:chung\)
\(→ΔABF\backsim ΔACE(g-g)\)
\(→\widehat{ABF}=\widehat{ACE}\) (2 góc tương ứng)
hay \(\widehat{EBK}=\widehat{FCK}\)
Xét \(ΔEKB\) và \(ΔFKC\):
\(\widehat{EBK}=\widehat{FCK}(cmt)\)
\(\widehat{EKB}=\widehat{FKC}\) (đối đỉnh)
\(→ΔEKB\backsim ΔFKC(g-g)\)
