Question

Cho \(\Delta ABC\) vuông tại \(A\) . Kẻ \(AH\perp BC\left(H\in BC\right)\) . Chứng minh rằng :

\(a.\) \(\frac{1}{AH^2}=\frac{1}{AB^2}+\frac{1}{AC^2}\)

\(b.\) \(AH^2=HB.HC\)

Answer 1

a. Ta có: \(\Delta ABC\) vuông tại \(A\)

\(\Rightarrow\) \(S_{\Delta ABC}=\frac{1}{2}AH.BC=\frac{1}{2}AB.AC\)

\(\Rightarrow AH.BC=AB.AC\)

\(\Rightarrow AH=\frac{AB.AC}{BC}\)

\(\Rightarrow\)\(\frac{1}{AH}=\frac{BC}{AB.AC}\)

\(\Rightarrow\)\(\frac{1}{AH^2}=\frac{BC^2}{AB^2.AC^2}\) (1)

Lại có: \(BC^2=AB^2+AC^2\) (định lý Pi-ta-go)

(1) \(\Rightarrow\) \(\frac{1}{AH^2}=\frac{AB^2+AC^2}{AB^2+AC^2}\)

\(\Rightarrow\) \(\frac{1}{AH^2}=\frac{1}{AB^2}+\frac{1}{AC^2}\) (đpcm)