Giải thích các bước giải:
a.Xét $\Delta ADC, \Delta ABE$ có:
$AD=AB$
$\widehat{DAC}=\widehat{DAB}+\widehat{BAC}=90^o+\widehat{BAC}=\widehat{EAC}+\widehat{BAC}=\widehat{EAB}$
$AC=AE$
$\to \Delta ADC=\Delta ABE(c.g.c)$
$\to \widehat{ADC}=\widehat{ABE}$
Gọi $AB\cap CD=F, CD\cap BE=G$
$\to \widehat{FDA}=\widehat{FBG}$
Mà $\widehat{DFA}=\widehat{BFG}$(đối đỉnh)
$\to \widehat{FGB}=180^o-\widehat{FBG}-\widehat{BFG}=180^o-\widehat{DFA}-\widehat{FDA}=\widehat{DAF}=90^o$
$\to BE\perp CD$