+ Vì \(EF\) // \(BC\left(gt\right)\)
=> \(\left\{{}\begin{matrix}\widehat{AEF}=\widehat{ABC}\\\widehat{AFE}=\widehat{ACB}\end{matrix}\right.\) (vì các góc đồng vị).
+ Xét 2 \(\Delta\) \(AEF\) và \(ABC\) có:
\(\widehat{AEF}=\widehat{ABC}\left(cmt\right)\)
\(\widehat{AFE}=\widehat{ACB}\left(cmt\right)\)
=> \(\Delta AEF\sim\Delta ABC\left(g-g\right).\)
=> \(\frac{AE}{AB}=\frac{AF}{AC}=\frac{EF}{BC}\) (các cặp cạnh tương ứng).
Mà \(\frac{AE}{AB}=\frac{2}{6}=\frac{1}{3}.\)
=> \(\frac{EF}{BC}=\frac{1}{3}.\)
=> \(\frac{EF}{10}=\frac{1}{3}\)
=> \(EF=\frac{1}{3}.10\)
=> \(EF=\frac{10}{3}\left(cm\right).\)
+ Xét \(\Delta AEF\) vuông tại \(A\left(gt\right)\) có:
\(AE^2+AF^2=EF^2\) (định lí Py - ta - go).
=> \(2^2+AF^2=\left(\frac{10}{3}\right)^2\)
=> \(AF^2=\left(\frac{10}{3}\right)^2-2^2\)
=> \(AF^2=\frac{100}{9}-4\)
=> \(AF^2=\frac{64}{9}\)
=> \(AF=\frac{8}{3}\left(cm\right)\) (vì \(AF>0\)).
Vậy \(AF=\frac{8}{3}\left(cm\right);EF=\frac{10}{3}\left(cm\right).\)
Chúc bạn học tốt!
