c/ Xét \(ΔBEA\) và \(ΔBAD\):
\(\widehat{EBA}\) hay \(\widehat{ABD}\): chung
\(\widehat{BEA}=\widehat{BAD}(=90^\circ)\)
\(→ΔBEA=ΔBAD(g-g)\)
\(→\dfrac{AB}{BE}=\dfrac{DB}{AB}\)
\(↔AB^2=BE.BD\)
Xét \(ΔBHA\) và \(ΔBAC\):
\(\widehat{ABH}\) hay \(\widehat{CBA}\): chung
\(\widehat{BHA}=\widehat{BAC}(=90^\circ)\)
\(→ΔBHA\backsim ΔBAC(g-g)\)
\(→\dfrac{BH}{BA}=\dfrac{BA}{BC}\)
\(↔AB^2=BH.BC\) mà \(AB^2=BE.BD\)
\(→BH.BC=BE.BD\)
d/ \(BH.BC=BE.BD↔\dfrac{BE}{BH}=\dfrac{BC}{BD}\)
Xét \(ΔBHE\) và \(ΔBAC\):
\(\widehat B\):chung
\(\dfrac{BE}{BH}=\dfrac{BC}{BD}\) (cmt)
\(→ΔBHE\backsim ΔBAC\)
\(→\widehat{BHE}=\widehat{BDC}\)
