a, Xét ΔABC và ΔHBA có :
\(\widehat{BAC}=\widehat{BHA}=90^0\)
\(\widehat{ABC}:chung\)
\(\Rightarrow\Delta ABC\sim\Delta HBA\left(g-g\right)\)
\(\Rightarrow\dfrac{AB}{BH}=\dfrac{BC}{AB}\Rightarrow AB^2=BH.BC\)
b, Xét ΔHBA và ΔHAC có :
\(\widehat{BHA}=\widehat{AHC}=90^0\)
\(\widehat{HBA}=\widehat{HAC}\) (cùng phụ \(\widehat{BAH}\) )
\(\Rightarrow\Delta HBA\sim\Delta HAC\left(g-g\right)\)
\(\Rightarrow\dfrac{AH}{HC}=\dfrac{BH}{AH}\Rightarrow AH^2=BH.HC\)
c, Dễ thấy :
+ \(\Delta BEH\sim\Delta HEA\left(g-g\right)\) ( vì \(\widehat{BEH}=\widehat{HEA}=90^0\) ; \(\widehat{EBH}=\widehat{EHA}\) do cùng phụ \(\widehat{EHB}\)) \(\Rightarrow\dfrac{BE}{EH}=\dfrac{EH}{AE}\Rightarrow EH^2=BE.AE\left(1\right)\)
+ \(\Delta AFH\sim\Delta HFC\left(g-g\right)\) ( vì \(\widehat{AFH}=\widehat{HFC}=90^0\); \(\widehat{FAH}=\widehat{FHC}\) do cùng phụ \(\widehat{AHF}\) ) \(\Rightarrow\dfrac{AF}{HF}=\dfrac{HF}{FC}\Rightarrow HF^2=AF.FC\left(2\right)\)
+ AEHF là hình chữ nhật vì \(\widehat{EAF}=\widehat{AEH}=\widehat{HFA}=90^0\) ⇒ AE = HF \(\left(3\right)\)
+ Áp dụng pythagore trong tam giác AEH vuông E : \(AH^2=HE^2+AE^2\left(4\right)\)
\(\left(1\right)\left(2\right)\left(3\right)\left(4\right)\Rightarrow AE.EB+AF.FC=AH^2\)
