1:
a: \(AH=\sqrt{2\cdot6}=2\sqrt{3}\left(cm\right)\)
\(AB=\sqrt{2\cdot8}=4\left(cm\right)\)
b: Xét ΔABC vuông tại A có sin C=AB/BC=1/2
nên góc C=30 độ
=>góc B=60 độ
2: \(\dfrac{BE}{CF}=\dfrac{BH^2}{AB}:\dfrac{CH^2}{AC}\)
\(=\dfrac{BH^2}{AB}\cdot\dfrac{AC}{CH^2}\)
\(=\left(\dfrac{BH}{CH}\right)^2\cdot\dfrac{AC}{AB}=\dfrac{AB^4}{AC^4}\cdot\dfrac{AC}{AB}=\dfrac{AB^3}{AC^3}\)