Ta có: \(\frac{1}{2}\widehat{A}+\widehat{B}+\widehat{ADB}\)\(=\frac{1}{2}\widehat{A}+\widehat{C}+\widehat{ADC}=180^o\)
mà \(\widehat{B}=\widehat{C}+20^o\)
=> \(\frac{1}{2}\widehat{A}+\widehat{C}+20^0+\widehat{ADB}=\frac{1}{2}\widehat{A}+\widehat{C}+\widehat{ADC}\)
\(\Rightarrow20^0+\widehat{ADB}=\widehat{ADC}\Rightarrow\widehat{ADC}-\widehat{ADB}=20^o\)(1)
Ta có: \(\widehat{ADB}+\widehat{ADC}=180^o\)(hai góc kề bù) (2)
Từ (1) và (2) suy ra: \(\widehat{ADC}=100^0;\widehat{ADB}=80^o\)