Giải thích các bước giải:

Xét $\Delta ABD,\Delta ACE$ có:

$AB=AE$

$\widehat{BAD}=90^o+\hat A=\widehat{CAE}$

$AC=AD$

$\to \Delta ABD=\Delta AEC(c.g.c)$

$\to \widehat{ABD}=\widehat{AEC}$

     $BD=CE$

Gọi $CE\cap BD=I, AB\cap CE=H$

$\to \widehat{AEH}=\widehat{HBI}$

$\to \widehat{HIB}=180^o-\widehat{IHB}-\widehat{IBH}=180^o-\widehat{AHE}-\widehat{HEA}=\widehat{HAE}=90^o$

$\to BD\perp CE$