Question

Answer 1

Ta có: \(\widehat{C_1}+\widehat{C_2}+\widehat{ACE} =360^o \)

\(\Rightarrow \widehat{C_1}+\widehat{C_1}-10^o +\widehat{C_1}-20^o=360^o \)

\(\Rightarrow 3 \widehat{C_1}-30^o=360^o \)

\(\Rightarrow \widehat{C_1}=130^o \)

\(\Rightarrow \widehat{C_2}=120^o \)

Ta có \(\widehat{BAC}+\widehat{C_1}=50^o+130^o=180^o \)

Mà hai góc ở vị trí bù nhau nên AB//CD

Lại có \(\widehat{C_2}+\widehat{CEF} =120^o+60^o=180^o \)

Mà hai góc ở vị trí bù nhau nên CD//EF (đpcm)