Giải thích các bước giải:
a.Ta có :
$\widehat{BAC}=180^o-\widehat{ABC}-\widehat{ACB}= 60^o$
$\to \widehat{DAC}=\widehat{BAD}=\dfrac 12 \widehat{BAC}=30^o$
$\to \widehat{ADC}=\widehat{BAD}+\widehat{ABD}=110^o$
b.Ta có $AB=AE,\widehat{DAB}=\widehat{DAE}\to\Delta ABD=\Delta AED(c.g.c)$
c.Từ câu b $\to\widehat{BDE}=2\widehat{ADB}=2(\widehat{DAC}+\widehat{ACD})=140^o$
Mà $\widehat{IBC}=\dfrac 12 \widehat{ABC}=40^o$
$\to\widehat{IBD}+\widehat{BDE}=180^o\to DE//BI$