Question

Cho hình thang vuông ABCD Góc B=C=90 độ. 2 đường chéo , AC vuông BC tạ H , AB= \(3\sqrt{5}\)cm AH= 3cm

a) Tính HB,HC,HD

b) \(\dfrac{1}{AB^2}-\dfrac{1}{CD^2}=\dfrac{1}{HB^2}-\dfrac{1}{HC^2}\)

Answer 1

a) \(AB^2=AH.AC=>\left(3\sqrt{5}\right)^2=3.AC=>AC=15=>BC=6\sqrt{5}\)
\(BH=\dfrac{AB.BC}{AC}=6\)
\(CH=AC-AH=12\)
\(HD.BH=CH^2=>HD.6=12^2=>HD=24\)
b) \(\left\{{}\begin{matrix}\dfrac{1}{CH^2}=\dfrac{1}{BC^2}+\dfrac{1}{CD^2}\\\dfrac{1}{BH^2}=\dfrac{1}{AB^2}+\dfrac{1}{BC^2}\end{matrix}\right.=>\dfrac{1}{AB^2}-\dfrac{1}{CD^2}=\dfrac{1}{HB^2}-\dfrac{1}{HC^2}\)