Giải thích các bước giải:
Sửa đề: $\widehat{CID}=60^o$
a.Ta có:
$\widehat{BAC}=180^o-(\hat B+\hat C)$
$\to \widehat{BAC}=180^o-2(\dfrac12\hat B+\dfrac12\hat C)$
$\to \widehat{BAC}=180^o-2(\widehat{IBC}+\widehat{ICB})$ vì $BI,CI$ là phân giác $\hat B,\hat C$
$\to \widehat{BAC}=180^o-2\widehat{CID}$
$\to \widehat{BAC}=180^o-2\cdot 60^o$
$\to \widehat{BAC}=60^o$
b.Ta có:
$\widehat{BEC}+\widehat{BDC}$
$=(180^o-\widehat{EBC}-\widehat{ECB})+(180^o-\widehat{DBC}-\widehat{DCB})$
$=(180^o-\hat B-\dfrac12\hat C)+(180^o-\dfrac12\hat B-\hat C)$
$=360^o-\dfrac32\hat B-\dfrac32\hat C$
$=360^o-\dfrac32(\hat B+\hat C)$
$=360^o-\dfrac32(180^o-\hat A)$
$=360^o-\dfrac32(180^o-60^o)$
$=180^o$
$\to \widehat{BEC},\widehat{BDC}$ bù nhau
