Ta có: $I$ là giao điểm $\widehat{B};\widehat{C}$

$⇒\widehat{IDA}=\widehat{IDC}=\dfrac{\widehat{B}}{2}$

$\widehat{IBA}=\widehat{IBC}=\dfrac{\widehat{D}}{2}$

$⇒\widehat{IDA}+\widehat{IBA}=\dfrac{\widehat{B}}{2}+\dfrac{\widehat{D}}{2}=\dfrac{\widehat{B}+\widehat{D}}{2}$

Mà tứ giác $ABCD$ ta có: $\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o$

$⇒\widehat{A}+\widehat{C}+\widehat{B}+\widehat{D}=360^o$

$⇒190^o+\widehat{B}+\widehat{D}=360^o$

$⇒\widehat{B}+\widehat{D}=170^o$
$⇒\widehat{IDA}+\widehat{IBA}=\dfrac{170^o}{2}=85^o$

Xét tứ giác $ABID$ có:

$\widehat{IDA}+\widehat{IBA}+\widehat{BID}+\widehat{A}=360^o$
$⇒85^o+145^o+\widehat{A}=360^o$

$⇒\widehat{A}=130^o$
$⇒\widehat{C}=190^o-\widehat{A}=190^o-130^o=60^o$
b4 Tên điểm trong hình

Ta có:$H$ là giao điểm $\widehat{B};\widehat{C}$
$⇒\widehat{HDC}=\dfrac{\widehat{D}}{2};\widehat{HCD}=\dfrac{\widehat{C}}{2}$

$⇒\widehat{HDC}+\widehat{HCD}=\dfrac{\widehat{D}}{2}+\dfrac{\widehat{C}}{2}=\dfrac{\widehat{D}+\widehat{C}}{2}$

$⇒\widehat{DHC}=180^o-(\widehat{HDC}+\widehat{HCD})=180^o-\dfrac{\widehat{D}+\widehat{C}}{2}$

Hay $\widehat{EHG}=180^o-(\widehat{HDC}+\widehat{HCD})=180^o-\dfrac{\widehat{D}+\widehat{C}}{2}$

Chứng minh tương tự như trên ta có:

$\widehat{EFG}=180^o-(\widehat{FAB}+\widehat{FBA})=180^o-\dfrac{\widehat{A}+\widehat{B}}{2}$

$⇒\widehat{EFG}+\widehat{EHG}=180^o-(\widehat{HDC}+\widehat{HCD})=180^o-\dfrac{\widehat{D}+\widehat{C}}{2}+180^o-\dfrac{\widehat{A}+\widehat{B}}{2}=360^o-\dfrac{\widehat{A}+\widehat{B}+\widehat{D}+\widehat{C}}{2}=360^o-\dfrac{360^o}{2}=180^o$
$⇒đpcm$