Question

Cho tam giác ABC cân tại A . Vẽ đường cao AH , BK . Chứng minh \(\dfrac{1}{BK^2}=\dfrac{1}{BC^2}+\dfrac{1}{4AH^2}\)

Answer 1

Xét \(\Delta ACH;\Delta BCK\) có

\(\left\{{}\begin{matrix}\widehat{C}\left(chung\right)\\\widehat{AHC}=\widehat{BKC}=90^o\end{matrix}\right.\)

\(\Rightarrow\Delta ACH\sim\Delta BCK\)

\(\Rightarrow\dfrac{AH}{BK}=\dfrac{CH}{CK}\)

\(\Rightarrow AH.CK=BK.CH\)

\(\Rightarrow AH^2.CK^2=BK^2.CH^2\)

\(\Rightarrow AH^2.CK^2=\dfrac{BK^2.BC^2}{4}\)

\(\Rightarrow AH^2.\left(BC^2-BK^2\right)=\dfrac{BK^2.BC^2}{4}\)

Chia cả 2 vế cho: \(AH^2.BC^2.BK^2\)

\(\Rightarrow\dfrac{1}{BK^2}-\dfrac{1}{BC^2}=\dfrac{1}{4AH^2}\)

\(\Rightarrow\dfrac{1}{BK^2}=\dfrac{1}{BC^2}+\dfrac{1}{4AH^2}\)