Question

Cho ΔABC vuông tại A có đường cao AH

a, CMR : BC = AH . cot_B + AH . cot_C

b, Kẻ HE ⊥ AB

CMR : BE = BC . cos³_B

c, Kẻ HF ⊥ AC

CMR : ΔAEF ~ ΔACB

d, CMR : \(\frac{BE}{CF}=\frac{AB^3}{AC^3}\)

e, \(\sqrt{\frac{BE}{AE}}=\frac{BH}{AH}\)

f, AH³ = BC . HE . HF

g, BE\(\sqrt{CH}\) + CF\(\sqrt{BH}\)= AH\(\sqrt{BC}\)

h, \(\sqrt[3]{BE^3}+\sqrt[3]{CF^3}=\sqrt[3]{BC^2}\)