a.
Theo giả thiết ta có \(\widehat{A}=\widehat{AEH}=\widehat{AFH}=90^0\Rightarrow AEHF\) là hcn
\(\Rightarrow EH=AF\Rightarrow EH^2+FH^2=AF^2+FH^2=AH^2\) (1)
Xét hai tam giác vuông AEH và HEB có:
\(\left\{{}\begin{matrix}\widehat{AEH}=\widehat{HEB}=90^0\\\widehat{EAH}=\widehat{EHB}\left(\text{cùng phụ }\widehat{B}\right)\end{matrix}\right.\)
\(\Rightarrow\Delta AEH\sim\Delta HEB\left(g.g\right)\)
\(\Rightarrow\dfrac{AE}{EH}=\dfrac{EH}{BE}\Rightarrow AE.BE=EH^2\) (2)
Tương tự, xét 2 tam giác vuông AFH và HFC có:
\(\left\{{}\begin{matrix}\widehat{AFH}=\widehat{HFC}=90^0\\\widehat{FAH}=\widehat{FHC}\left(\text{cùng phụ }\widehat{C}\right)\end{matrix}\right.\)
\(\Rightarrow\Delta AFH\sim\Delta HFC\Rightarrow\dfrac{AF}{HF}=\dfrac{HF}{CF}\Rightarrow AF.CF=HF^2\) (3)
(1);(2);(3) \(\Rightarrow AE.BE+AF.CF=EH^2+FH^2=AH^2\)
b.
Trong tam giác vuông ABC: \(cos\widehat{B}=\dfrac{AB}{BC}\)
Trong tam giác vuông BEH: \(cos\widehat{B}=\dfrac{BE}{BH}\)
Trong tam giác vuông ABH: \(cos\widehat{B}=\dfrac{BH}{AB}\)
\(\Rightarrow cos\widehat{B}.cos\widehat{B}.cos\widehat{B}=\dfrac{AB}{BC}.\dfrac{BE}{BH}.\dfrac{BH}{AB}\)
\(\Rightarrow cos^3\widehat{B}=\dfrac{BE}{BC}\)
\(\Rightarrow BE=BC.cos^3\widehat{B}\)
