Giải thích các bước giải:
Ta có $AE$ là phân giác trong góc $A, AF$ là phân giác ngoài góc $A\to AE\perp AF$
Tương tự $BE\perp BF$
Ta có:
$\begin{split}\widehat{AEB}&=180^o-\widehat{BAE}-\widehat{ABE}\\&=180^o-\dfrac12\widehat{BAD}-\dfrac12\widehat{ABC}\\&=\dfrac12\cdot (360^o-\widehat{BAD}-\widehat{ABC})\\&=\dfrac12(\widehat{ADC}+\widehat{BCD})\\&=\dfrac12(\hat C+\hat D)\end{split}$
Ta có $\widehat{AFB}+\widehat{FAE}+\widehat{AEB}+\widehat{EBF}=360^o$
$\to\widehat{AFB}+90^o+\widehat{AEB}+90^o=360^o$
$\to\widehat{AFB}=180^o-\widehat{AEB}$
$\to\widehat{AFB}=180^o-\dfrac12(\hat C+\hat D)$
$\to\widehat{AFB}=\dfrac12(360^o-\hat C-\hat D)$
$\to\widehat{AFB}=\dfrac12(\hat B+\hat A)$
