Giải thích các bước giải:
a.Ta có:
$\widehat{DBC}=\widehat{ABE}=90^o$
$\to \widehat{DBC}-\widehat{ABC}=\widehat{ABE}-\widehat{ABC}$
$\to\widehat{ABD}=\widehat{EBC}$
Mà $BD=BC, BA=BE$
$\to \Delta BAD=\Delta BEC(c.g.c)$
$\to DA=EC$
b.Gọi $AD\cap CE=F$
Từ câu a $\to\widehat{DAB}=\widehat{BEC}$
Ta có:
$\widehat{FAC}+\widehat{FCA}=(180^o-\widehat{DAB}-\widehat{BAC})+(180^o-\widehat{ACB}-\widehat{BCE})$
$\to \widehat{FAC}+\widehat{FCA}=180^o-\widehat{DAB}-\widehat{BAC}+180^o-\widehat{ACB}-\widehat{BCE}$
$\to \widehat{FAC}+\widehat{FCA}=360^o-(\widehat{DAB}+\widehat{BCE})-(\widehat{BAC}+\widehat{ACB})$
$\to \widehat{FAC}+\widehat{FCA}=360^o-(\widehat{DAB}+\widehat{DAB})-(180^o-\widehat{ABC})$
$\to \widehat{FAC}+\widehat{FCA}=360^o-(180^o-\widehat{DBA})-(180^o-\widehat{ABC})$
$\to \widehat{FAC}+\widehat{FCA}=360^o-180^o+\widehat{DBA}-180^o+\widehat{ABC}$
$\to \widehat{FAC}+\widehat{FCA}=\widehat{DBA}+\widehat{ABC}$
$\to \widehat{FAC}+\widehat{FCA}=\widehat{DBC}$
$\to \widehat{FAC}+\widehat{FCA}=90^o$
$\to \Delta FAC$ vuông tại $F$
$\to FA\perp FC$
$\to AD\perp CE$
