Question

cho △ABC⊥A, đường cao AH, D và E lần lượt là hình chiếu của H trên AB, AC. chứng minh

a)\(\dfrac{HB}{HC}=\left(\dfrac{AB}{AC}\right)^2\)

b)\(\dfrac{CE}{BD}=\left(\dfrac{CA}{AB}\right)^3\)

c)\(AH^3=BC.BD.CE\)

d)\(3AH^2+BD^2+CE^2=BC^2\)

lm nhanh giúp mk nhé! Mk đang càn gấp lắm!

Answer 1

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: \(3AH^2+BD^2+CE^2=2AH^2+\left(DH^2+BD\right)^2+\left(HE^2+CE^2\right)\)

\(=2.HB.HC+BH^2+CH^2=\left(BH+CH\right)^2=BC^2\)