Đáp án + Giải thích các bước giải:

Xét tứ giác $ABCD$, ta có:

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

$\Rightarrow \widehat{A}+ \widehat{B} + \widehat{C} + 2\widehat{D} = 360^\circ + \widehat{D}$

$\Rightarrow \widehat{A} +\widehat{C} + 180^\circ = 360^\circ + \widehat{D}$

$\Rightarrow \widehat{A} + \widehat{C} = 180^\circ +\widehat{D}$

$\Rightarrow \widehat{BAD} + \widehat{BCD} = 180^\circ + \widehat{ADC}$

Xét tứ giác $AICB$, ta có:

$\widehat{IAB} + \widehat{ABC} + \widehat{BCI} + \widehat{AIC} = 360^\circ$

$\Rightarrow \dfrac{1}{2}\widehat{BAD} + \widehat{ABC} + \dfrac{1}{2}\widehat{BCD} + 150^\circ = 360^\circ$

$\Rightarrow \dfrac{1}{2}\left(\widehat{BAD} + \widehat{BCD}\right) + \widehat{ABC} = 210^\circ$

$\Rightarrow 90^\circ + \dfrac{1}{2}\widehat{ADC} + \widehat{ABC} = 210^\circ$

$\Rightarrow \dfrac{1}{2}\widehat{ADC} + \widehat{ABC} = 120^\circ$

Mà $\widehat{ABC} +2\widehat{ADC} = \widehat{ABC} + \dfrac{1}{2}\widehat{ADC} + \dfrac{3}{2}\widehat{ADC} = 180^\circ$

$\Rightarrow \dfrac{3}{2}\widehat{ADC} = 60^\circ$

$\Rightarrow \widehat{ADC} = 40^\circ$

$\Rightarrow \widehat{ABC} = 180^\circ - 2\widehat{ADC} = 180^\circ - 80^\circ = 100^\circ$

$\phantom{\Rightarrow} \widehat{BAD} + \widehat{BCD} = 180^\circ + \widehat{ADC} = 180^\circ + 40^\circ = 220^\circ$

Giả sử đặt $\widehat{BAD} = x^\circ$ thì $\widehat{BCD} = 220^\circ - x^\circ$

Vậy $\widehat{BAD} = x^\circ, \widehat{ABC} = 100^\circ, \widehat{BCD} = 220^\circ - x^\circ, \widehat{ADC} = 40^\circ$