Cho $\Delta ABC;\widehat{A}=90^{\circ},AH\perp BC,HE\perp AB,HF\perp AC$.CMR
a)$AE.AB=AF.AC$
b)$\Delta AEF\sim\Delta ACB$
c)$BC^{2}=3AH^{2}+BE^{2}+CF^{2}$
d)$\frac{AB^{3}}{AC^{3}}=\frac{BE}{CF}$
e)$AH^{3}=BC.BE.CF$
f)$BE.\sqrt{CH}+CE.\sqrt{BH}=AH.\sqrt{BC}$
g)$\sqrt[3]{BE^{2}}+\sqrt[3]{CF^{2}}=\sqrt[3]{BC^{2}}$
ps:câu a,b mình làm được rồi
