Giúp mình vs mình cần gấp ạ

4) 
Ta có $\widehat{ADB} + \widehat{ADC} = 180^\circ$ (hai góc kề bù).
Trong $\triangle ABD$: $\widehat{ADB} = 180^\circ - \widehat{B} - \widehat{BAD}$.
Trong $\triangle ACD$: $\widehat{ADC} = 180^\circ - \widehat{C} - \widehat{CAD}$.
Vì AD là phân giác $\widehat{BAC}$ nên $\widehat{BAD} = \widehat{CAD}$.
=>

$\widehat{ADB} - \widehat{ADC} = (180^\circ - \widehat{B} - \widehat{BAD}) - (180^\circ - \widehat{C} - \widehat{CAD}) = \widehat{C} - \widehat{B}$.
Mà $\widehat{B} - \widehat{C} = 40^\circ \Rightarrow \widehat{C} - \widehat{B} = -40^\circ$.
Vậy $\widehat{ADB} - \widehat{ADC} = -40^\circ$.
Ta có hệ:
$\begin{cases} \widehat{ADB} + \widehat{ADC} = 180^\circ \\ \widehat{ADB} - \widehat{ADC} = -40^\circ \end{cases}$
Cộng vế với vế: $2\widehat{ADB} = 140^\circ \Leftrightarrow \widehat{ADB} = 70^\circ$.
Trừ vế với vế: $2\widehat{ADC} = 220^\circ \Leftrightarrow \widehat{ADC} = 110^\circ$.
Đáp số: $\widehat{ADB} = 70^\circ$, $\widehat{ADC} = 110^\circ$.

5.
Xét $\triangle EBC$: $\widehat{BEC} = 180^\circ - (\widehat{EBC} + \widehat{ECB})$.
Xét $\triangle ABC$: $\widehat{BAC} = 180^\circ - (\widehat{ABC} + \widehat{ACB})$.
Mà $\widehat{ABC} = \widehat{ABE} + \widehat{EBC}$ và $\widehat{ACB} = \widehat{ACE} + \widehat{ECB}$.
Thay vào biểu thức của $\widehat{BAC}$:
$\widehat{BAC} = 180^\circ - (\widehat{ABE} + \widehat{EBC} + \widehat{ACE} + \widehat{ECB})$
$\widehat{BAC} = 180^\circ - (\widehat{EBC} + \widehat{ECB}) - (\widehat{ABE} + \widehat{ACE})$.
Thay $180^\circ - (\widehat{EBC} + \widehat{ECB}) = \widehat{BEC}$ vào:
$\widehat{BAC} = \widehat{BEC} - (\widehat{ABE} + \widehat{ACE})$
$\widehat{BEC} = \widehat{ABE} + \widehat{ACE} + \widehat{BAC}$.