Giải thích các bước giải:
2.Xét $\Delta CDG, \Delta ECB$ có:
$CD=AB=BE$
$\widehat{GDC}=90^o+\widehat{ADC}=90^o+\widehat{ABC}=\widehat{EBC}$
$GD=AD=BC$
$\to\Delta CDG=\Delta EBC(c.g.c)$
$\to CG=CE,\widehat{DCG}=\widehat{CEB}$
$\to \widehat{GCE}=\widehat{DCB}-\widehat{GCD}-\widehat{ECB}$
$\to \widehat{GCE}=\widehat{DCB}-\widehat{CEB}-\widehat{ECB}$
$\to \widehat{GCE}=\widehat{DCB}-(\widehat{CEB}+\widehat{ECB})$
$\to \widehat{GCE}=\widehat{DCB}-(180^o-\widehat{EBC})$
$\to \widehat{GCE}=\widehat{DCB}-180^o+\widehat{EBC}$
$\to \widehat{GCE}=\widehat{DCB}-180^o+\widehat{EBA}+\widehat{ABC}$
$\to \widehat{GCE}=\widehat{DCB}+\widehat{ABC}-180^o+\widehat{EBA}$
$\to \widehat{GCE}=180^o-180^o+\widehat{EBA}$ vì $AB//CD$
$\to \widehat{GCE}=\widehat{EBA}$
$\to \widehat{GCE}=90^o$
Kết hợp $GC=CE\to\Delta CGE$ vuông cân tại C$