Giả sử $AC$ là đường chéo lớn của hình bình hành $ABCD$  ,từ $C$ kẻ đường vuông góc $CE$ với đường thẳng $AB$ đường vuông góc $CF$ vs đường thẳng $AD$.Chứng minh rằng $AB.AE+ AD.AF=AC^{2}$
$AC^2=AF^2+CF^2=AD^2+DF^2+FC^2+2AD.DF$
$=AD^2+DC^2+2AD.DF.$
$AB.AE+AD.AF=DC.(AB+BE)+BC(AD+DF)$
$=DC.AB+DC.BE+BC.AD+BC.DF=DC^2+AD^2+2AD.DF.$
(Vì $\triangle CFD\sim \triangle CEB$ $(g.g)\Rightarrow \frac{DC}{DF}=\frac{BC}{BE}\Rightarrow DC.BE=BC.DF$ )
Vậy $AC^2=AB.AE+AD.AF$.
h đã lm đc r –  Yêu Tatoo 12-03-16 07:37 PM
  • Từ B, D kẻ đường vuông góc xuống AC cắt AC tại H,K
  • ta có tg ADH đồng dạng ACF
  • => AD.AF=AH.AC
  •  tương tự có AB.AE=AK.AC
  •  => AB.AE+ AD.AF=AC(AH+AK)=AC^2
r e vẽ xg r c –  Yêu Tatoo 12-03-16 07:40 PM
thử vẽ lại hình đi e. AC là đường chéo lớn nhé –  rang 12-03-16 07:39 PM
hay e vẽ hình sai nnhir –  Yêu Tatoo 12-03-16 07:36 PM
theo e vẽ hình thì đ H fai trùng vs đ C thì ms có tam giác ADH –  Yêu Tatoo 12-03-16 07:36 PM

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