Lời giải:
Theo định lý Talet:
\(\frac{AE}{EF}=\frac{AB}{CF}\Rightarrow \frac{AE}{AF}=\frac{AB}{AB+CF}=\frac{AB}{DC+CF}=\frac{AB}{DF}\)
\(\Rightarrow AE=\frac{AB.AF}{DF}\)
Do đó:
\(\frac{1}{AE^2}+\frac{1}{AF^2}=\frac{DF^2}{AB^2AF^2}+\frac{1}{AF^2}=\frac{1}{AF^2}.\frac{DF^2+AB^2}{AB^2}\)
\(=\frac{1}{AF^2}.\frac{DF^2+AD^2}{AB^2}=\frac{1}{AF^2}.\frac{AF^2}{AB^2}=\frac{1}{AB^2}\)
(đpcm)
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