Question

Cho hình thoi ABCD có góc B tù và BH là đường cao. Chứng minh rằng 

\(\dfrac{1}{BH^2}=\dfrac{1}{AC^2}+\dfrac{1}{BD^2}\)

Answer 1

Giả sử \(BH\perp AD\)

Gọi  \(O=AC\cap BD\)

Có \(S_{ABCD}=\dfrac{1}{2}AC.BD=BH.AD\)

\(\Leftrightarrow\left\{{}\begin{matrix}AC.BD=2S_{ABCD}\\BH=\dfrac{S_{ABCD}}{AD}\end{matrix}\right.\)

Có \(\dfrac{1}{AC^2}+\dfrac{1}{BD^2}=\dfrac{AC^2+BD^2}{AC^2.BD^2}=\dfrac{4\left(OA^2+OD^2\right)}{\left(2S_{ABCD}\right)^2}\)\(=\dfrac{4AD^2}{4S_{ABCD}}=\dfrac{1}{BH^2}\) 

Vậy \(\dfrac{1}{BH^2}=\dfrac{1}{AC^2}+\dfrac{1}{BD^2}\)