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a: \(AB^3:BD=AB^3:\dfrac{BH^2}{AB}=AB^3\cdot\dfrac{AB}{BH^2}\)
\(=\dfrac{AB^4}{BH^2}=\left(\dfrac{AB^2}{BH}\right)^2=BC^2\)
=>\(BC^2\cdot BD=AB^3\)
\(\dfrac{AC^3}{CE}=AC^3:\dfrac{CH^2}{AC}=\dfrac{AC^4}{CH}=BC^2\)
=>\(BC^2\cdot AE=AC^3\)
b: \(BC\cdot BD\cdot CE=BC\cdot\dfrac{BH^2}{AB}\cdot\dfrac{CH^2}{AC}\)
\(=\dfrac{AH^4}{AH}=AH^3\)
góc ADH=góc AEH=góc DAE=90 độ
=>ADHE là hình chữ nhật
=>AH=DE
BD*CE*BC
=BH^2/BA*CH^2/CA*BC
=AH^4/AH=AH^3
=DE^3
a) Ta có: \(\left(\dfrac{AB}{AC}\right)^2=\dfrac{AB^2}{AC^2}=\dfrac{BH.BC}{CH.BC}=\dfrac{BH}{HC}\)
b) Ta có: \(\left(\dfrac{CA}{AB}\right)^4=\left(\dfrac{CA^2}{AB^2}\right)^2=\left(\dfrac{CH.BC}{BH.BC}\right)^2=\dfrac{CH^2}{BH^2}=\dfrac{CE.CA}{BD.BA}\)
\(=\dfrac{CE}{BD}.\dfrac{CA}{BA}\Rightarrow\left(\dfrac{CA}{AB}\right)^3=\dfrac{CE}{BD}\)
c) Ta có: \(AH^4=\left(AH^2\right)^2=\left(BH.CH\right)^2=BH^2.CH^2\)
\(=BD.BA.CE.CA=BD.CE\left(AB.AC\right)=BD.CE.AH.BC\)
\(\Rightarrow BD.CE.BC=AH^3\)
d) Vì \(\angle HDA=\angle HEA=\angle DAE=90\Rightarrow ADHE\) là hình chữ nhật
\(\Rightarrow AH=DE\Rightarrow AH^2=DE^2=DH^2+HE^2\)
Ta có: \(3AH^2+BD^2+CE^2=2AH^2+\left(DH^2+BD\right)^2+\left(HE^2+CE^2\right)\)
\(=2.HB.HC+BH^2+CH^2=\left(BH+CH\right)^2=BC^2\)
a: \(AB^3:BD=AB^3:\dfrac{BH^2}{AB}=AB^3\cdot\dfrac{AB}{BH^2}\)
\(=\dfrac{AB^4}{BH^2}=\left(\dfrac{AB^2}{BH}\right)^2=BC^2\)
=>\(BC^2\cdot BD=AB^3\)
\(\dfrac{AC^3}{CE}=AC^3:\dfrac{CH^2}{AC}=\dfrac{AC^4}{CH}=BC^2\)
=>\(BC^2\cdot AE=AC^3\)
b: \(BC\cdot BD\cdot CE=BC\cdot\dfrac{BH^2}{AB}\cdot\dfrac{CH^2}{AC}\)
\(=\dfrac{AH^4}{AH}=AH^3\)