a) *Xét ΔEFD và ΔFDB có:
\(\left\{{}\begin{matrix}DF.l\text{à}.c\text{ạnh}.chung\\\widehat{BDF}=\widehat{EFD}\left(2.g\text{óc}.so.le.trong.do.AB//EF\right)\\\widehat{BFD}=\widehat{EDF}\left(2.g\text{óc}.so.le.trong.do.DE//BC\right)\end{matrix}\right.\)
\(\Rightarrow\Delta EFD=\Delta FBD\left(g-c-g\right)\)
⇒ BD = EF (hai góc tương ứng)
Mà \(\left\{{}\begin{matrix}AD=BD\left(gt\right)\\BD=EF\left(cmt\right)\end{matrix}\right.\)
⇒ AD = EF
b) *Ta có: \(\left\{{}\begin{matrix}\widehat{ADE}=\widehat{DBF}\left(2.g\text{óc}.so.le.trong.do.DE//BC\right)\\\widehat{DBF}=\widehat{EFC}\left(2.g\text{óc}.so.le.trong.do.AB//EF\right)\end{matrix}\right.\)
⇒ \(\widehat{ADE}=\widehat{EFC}\)
*Xét \(\Delta ADE\) và \(\Delta EFC\) có:
\(\left\{{}\begin{matrix}AD=EF\left(cmt\right)\\\widehat{ADE}=\widehat{EFC\left(cmt\right)}\\\widehat{DAF}=\widehat{FEC}\left(\text{đ}\text{ồng}.v\text{ị}.do.AB//EF\right)\end{matrix}\right.\)
⇒ ΔADE = ΔEFC (g-c-g)
