a) \(\dfrac{AB^2}{AC^2}=\dfrac{BH.BC}{CH.BC}=\dfrac{BH}{CH}\)
b) Vì \(\angle HDA=\angle HEA=\angle DAE=90\Rightarrow ADHE\) là hình chữ nhật
\(\Rightarrow DE=AH\)
Ta có: \(AH^4=\left(AH^2\right)^2=\left(BH.CH\right)^2=BH^2.CH^2\)
\(=BD.BA.CE.CA=BD.CE.AH.BC\left(AB.AC=AH.BC\right)\)
\(\Rightarrow AH^3=BD.CE.BC\)


