Ta có: $\widehat{B} + \widehat{C} = 200^o$
$\widehat{B} + \widehat{D} = 180^o$
⇒ $\widehat{C} - \widehat{D} = 20^o$
mà $\widehat{C} + \widehat{D} = 120^o$
nên $2\widehat{C} = 140^o$
⇒ $\widehat{C} = 70^o$
⇒ $\widehat{D} = 120 - 70 = 50^o$
⇒ $\widehat{B} = 200 - 70 = 130^o$
⇒ $\widehat{A} = 360^o - (\widehat{B} +\widehat{C} + \widehat{D}) = 360 - (70 + 50 + 130) = 110^o$
Ta có: $\widehat{AIB} = 180^o - (\widehat{IAB} + \widehat{IBA})$
$=180^o - (\dfrac{\widehat{DAB}}{2} + \dfrac{\widehat{CBA}}{2})$
$= \dfrac{360^o - (\widehat{A} + \widehat{B})}{2}$
$= \dfrac{\widehat{C} + \widehat{D}}{2}$
⇒ $2\widehat{AIB} = \widehat{C} + \widehat{D}$
