+ AB // CD \(\Rightarrow\left\{{}\begin{matrix}\widehat{BAK}=\widehat{AKD}\\\widehat{ABK}=\widehat{BKC}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\widehat{DAK}=\widehat{AKD}\left(1\right)\\\widehat{CBK}=\widehat{BKC}\end{matrix}\right.\)
(1) => ΔADK cân tại D
=> AD = DK
+ tương tự : BC = CK
Do đó : AD + BC = DK + CK = CD