a/ Xét $\Delta{ABC}$ và $\Delta{ADE}$:
$\widehat{BAC}=\widehat{DAE}(=90^o)$
$AB=AD(gt)$
$AC=AB(gt)$
$\to \Delta{ABC}=\Delta{ADE}(c-g-c)$
b/ Gọi $ED∩BC≡F$
$\Delta{ABC}=\Delta{ADE}$
$\to \widehat{E}=\widehat{C}$
Xét $\Delta{ADE}$:
$\widehat{E}+\widehat{ADE}=90^o$
mà $\widehat{E}=\widehat{C}$ và $\widehat{ADE}=\widehat{FDC}$
$\to \widehat{C}+\widehat{FDC}=90^o$
$\to \Delta{FDC}$ vuông tại $F$
$\to DE\perp BC$