Giải Bài tập 4 trang 67 sgk Toán 8 tập 1 Chân trời

Ta có: $\widehat{BAD}+\widehat{A}_{ngoài}=180^{\circ}$ (hai góc kề bù)

Do đó: $\widehat{BAD}+65^{\circ}=180^{\circ}\Rightarrow \widehat{BAD}=180^{\circ}-65^{\circ}=115^{\circ}$

$\widehat{ABC}+\widehat{B}_{ngoài}=180^{\circ}$ (hai góc kề bù)

Do đó: $\widehat{ABC}+100^{\circ}=180^{\circ}\Rightarrow \widehat{ABC}=180^{\circ}-100^{\circ}=80^{\circ}$

$\widehat{BCD}+\widehat{C}_{ngoài}=180^{\circ}$ (hai góc kề bù)

Do đó: $\widehat{BCD}+60^{\circ}=180^{\circ}\Rightarrow \widehat{BCD}=180^{\circ}-60^{\circ}=120^{\circ}$

Tứ giác ABCD có $\widehat{BAD}+\widehat{ABC}+\widehat{BCD}+\widehat{ADC}=360$

Do đó: $115^{\circ}+80^{\circ}+120^{\circ}+\widehat{ADC}=360^{\circ}\Rightarrow \widehat{ADC}=360^{\circ}-(115^{\circ}+80^{\circ}+120^{\circ})=45^{\circ}$

Ta có $\widehat{D}_{ngoài}+\widehat{ADC}=180^{\circ}$ (hai góc kề bù)

Do đó $\widehat{D}_{ngoài}+45^{\circ}=180^{\circ}\Rightarrow \widehat{D}_{ngoài}=180^{\circ}-45^{\circ}=135^{\circ}$

Vậy góc ngoài tại đỉnh D bằng $135^{\circ}$