Cho tam giác ABC vuông ở A, có đường cao AH.
a)Cm \(\frac{AB^2}{AC^2}=\frac{HB}{HC}\)
b) vẽ HE\(\perp\)AB, HF\(\perp\)AC. cm \(\frac{AB^3}{AC^3}=\frac{BE}{CF}\)
c) cm \(\Delta AEF\)đồng dạng \(\Delta ABC\)
d)CM \(BC^2=3AH^2+BE^2+CF^2\)
e)CM \(AH^3=BC.BE.CF\)
f)CM \(BE\sqrt{CH}+CF\sqrt{BH}=AH\sqrt{BC}\)
g)CM \(^{\sqrt[3]{BE^2}+\sqrt[3]{CF^2}=\sqrt[3]{BC^2}}\)