Giải thích các bước giải:

a.Ta có:

$\hat C+\hat D=360^o-\hat B-\hat A=150^o$ 

$\to \dfrac12(\hat C+\hat D)=75^o$ 

$\to \dfrac12\hat C+\dfrac12\hat D=75^o$ 

$\to\widehat{OCD}+\widehat{ODC}=75^o$

$\to 180^o-(\widehat{OCD}+\widehat{ODC})=105^o$

$\to\widehat{COD}=105^o$

b.Ta có:

$\widehat{COD}=180^o-(\widehat{OCD}+\widehat{OCD})$

$\to \widehat{COD}=180^o-(\dfrac12\widehat{ADC}+\dfrac12\widehat{DCB})$

$\to \widehat{COD}=180^o-\dfrac12(\widehat{ADC}+\widehat{DCB})$

$\to \widehat{COD}=180^o-\dfrac12(360^o-(\widehat{DAB}+\widehat{CBA}))$

$\to \widehat{COD}=\dfrac12(\widehat{DAB}+\widehat{CBA})$

$\to \widehat{COD}=\dfrac12(\widehat{A}+\widehat{B})$

c.Từ câu b

$\to \widehat{AIB}=\dfrac12(\hat D+\hat C)$

$\to \widehat{FIE}=\dfrac12(\hat D+\hat C)$

Mà $\widehat{FOE}=\widehat{COD}=\dfrac12(\widehat{A}+\widehat{B})$

$\to \widehat{FIE}+\widehat{FOE}=\dfrac12(\hat D+\hat C)+\dfrac12(\widehat{A}+\widehat{B})$

$\to \widehat{FIE}+\widehat{FOE}=\dfrac12(\hat D+\hat C+\widehat{A}+\widehat{B})$

$\to \widehat{FIE}+\widehat{FOE}=180^o$

$\to \widehat{IFO}+\widehat{IEO}=360^o-(\widehat{FIE}+\widehat{FOE})$

$\to \widehat{IFO}+\widehat{IEO}=180^o$