Giải thích các bước giải:
Ta có : $AB=AC, AD=AB ,CB=CE$
$\to\widehat{BAC}=\widehat{BCA}=90^o-\dfrac 12\widehat{ABC}=40^o$
$\to\widehat{ABD}=\widehat{ADB}=90^o-\dfrac 12\widehat{BAD}=70^o$
Tương tự $\widehat{BED}=\widehat{EBC}=90^o-\dfrac 12\widehat{BCE}=70^o$
$\to \widehat{EBD}=180^o-\widehat{BEC}-\widehat{BDA}=40^o$