a/ Xét \(ΔABH\) và \(ΔCBA\):

\(\widehat B:chung\)

\(\widehat{AHB}=\widehat{CAB}(=90^\circ)\)

\(→ΔABH\backsim ΔCBA(g-g)\)

\(→\dfrac{AB}{BH}=\dfrac{BC}{AB}\)

\(↔AB^2=BH.BC\)

b/ Xét \(ΔABD\) và \(ΔHBK\):

\(\widehat{ABD}=\widehat{HBK}\) (\(BD\) hay \(BK\) là đường phân giác \(\widehat B\))

\(\widehat{BAD}=\widehat{BHK}(=90^\circ)\)

\(→ΔABD\backsim ΔHBK(g-g)\) 

\(AB^2=BH.BC↔BH=\dfrac{AB^2}{BC}=\dfrac{4^2}{6}=\dfrac{8}{3}(cm)\)

\(\dfrac{S_{ΔABD}}{S_{ΔHBK}}=(\dfrac{BA}{BH})^2=(\dfrac{4}{\dfrac{8}{3}})^2=\dfrac{9}{4}\)

c/ \(\widehat{AKD}=\widehat{BKH}\) (đối đỉnh)

\(ΔABD\backsim ΔHBK→\widehat{ADB}=\widehat{BKH}\) hay \(\widehat{ADK}=\widehat{BKH}\)

\(→\widehat{AKD}=\widehat{ADK}\)

\(→ΔAKD\) cân tại \(A\)

d/ \(BK\) là đường phân giác \(\widehat B\)

\(→\dfrac{AK}{HK}=\dfrac{BA}{BH}\) (1)

\(BD\) là đường phân giác \(\widehat B\)

\(→\dfrac{CD}{AD}=\dfrac{BC}{BA}\) (2)

Lại có: \(\dfrac{BA}{BH}=\dfrac{BC}{BA}\) (2)

(1)(2)(3) \(→\dfrac{AK}{HK}=\dfrac{CD}{AD}\)

\(↔HK.CD=AK.AD\)