$DE=EC\Rightarrow \triangle DEC$ cân $\Rightarrow \widehat{EDC}=\widehat{ECD}$$\Rightarrow \widehat{EDA}=\widehat{ECB}$ mà $AD=BC;DE=EC$$\Rightarrow \triangle ADE=\triangle BCE$(c-g-c)$\Rightarrow AE=BE....$
TH:BH cắt AD tại M$DE=EC\Rightarrow \triangle DEC$ cân $\Rightarrow \widehat{EDC}=\widehat{ECD}$$\Rightarrow \widehat{EDA}=\widehat{ECB}$ mà $AD=BC;DE=EC$$\Rightarrow \triangle ADE=\triangle BCE$(c-g-c)$\Rightarrow AE=BE....$