Question

cho ΔABC vuông tại A có đường cao AH. Chứng minh:

a) AB² = BC . BH; AC² = BC . CH. Từ đó cm định lí pytago

b) AH² = BH . CH

c) \(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\)

d) AH . BC = AB . AC

Answer 1

a) Xét ΔABH và ΔABC có 
\(\widehat{AHB}=\widehat{BAC}=90^o\)
\(\widehat{B}\) chung 
=> ΔABH ∼ ΔABC (g.g) \(=>\dfrac{AB}{BH}=\dfrac{BC}{AB}=>AB^2=BH.BC\)
Xét ΔACH và ΔABC có 
\(\widehat{AHC}=\widehat{BAC}=90^o\)
\(\widehat{C}\) chung
=> ΔACH ∼ ΔABC (g.g) \(=>\dfrac{AC}{CH}=\dfrac{BC}{AC}=>AC^2=CH.BC\)

b) Xét ΔABH và ΔACH có \(\widehat{AHB}=\widehat{AHC}=90^o\) ; \(\widehat{BAH}=\widehat{ACH}\)
=> ΔABH ∼ ΔACH (g.g) \(=>\dfrac{AH}{BH}=\dfrac{CH}{AH}=>AH^2=BH.CH\)
c) Vì ΔABC vuông tại A \(=>S_{ABC}=\dfrac{AH.BC}{2}=\dfrac{AB.AC}{2}=>AH.BC=AB.AC\)
\(=>AH=\dfrac{AB.AC}{BC}=>\dfrac{1}{AH}=\dfrac{BC}{AB.AC}=>\dfrac{1}{AH^2}=\dfrac{BC^2}{AB^2.AC^2}\)
d) Vì ΔABH ∼ ΔABC \(=>\dfrac{AH}{AB}=\dfrac{AC}{BC}=>AH.BC=AB.AC\)