Ta có : AE = CD
=> AE + EC = CD + EC = AC = ED
Ta lại có : \(\widehat{E1}+\widehat{E2}=\widehat{C1}+\widehat{C1}=180^o\)
Mà \(\widehat{E1}=\widehat{C1}\)
\(\Rightarrow\widehat{E2}=\widehat{C2}\)
- Xét tam ABC và tam giác DFE có
\(\widehat{E2}=\widehat{C2}\)
AC = ED
\(\widehat{A}=\widehat{D}\)
=> Tam giác ABC = Tam giác DFE
=> AB = DE .
Ta có: \(\left\{{}\begin{matrix}\widehat{FED}=180^0-\widehat{E_1}\\\widehat{ACB}=180^0-\widehat{C_1}\end{matrix}\right.\)
Mà \(\widehat{E_1}=\widehat{C_1}\) \(\Rightarrow\widehat{FED}=\widehat{ACB}\)
Xét hai tam giác ABC và tam giác DFE có:
\(\left\{{}\begin{matrix}\widehat{BAC}=\widehat{FDE}\\AC=DE\left(AE=CD\right)\\\widehat{ACB}=\widehat{FED}\end{matrix}\right.\)
\(\Rightarrow\Delta ABC=\Delta DFE\) (g-c-g)
\(\Rightarrow AB=DF\)