\(\cdot\text{Do Ax//Cz}\)
\(\Rightarrow\widehat{xAC}+\widehat{ACz}=180^o\left(\text{trong cùng phía}\right)\left(1\right)\)
\(\text{Thay }\widehat{xAC}=120^o\text{ vào }\left(1\right)\)
\(\Rightarrow120^o+\widehat{ACz}=180^o\)
\(\Rightarrow\widehat{ACz}=180^o-120^o\)
\(\Rightarrow\widehat{ACz}=60^o\)
\(\cdot\text{Do By//Cz}\)
\(\Rightarrow\widehat{yCB}+\widehat{BCz}=180^o\left(\text{trong cùng phía}\right)\left(2\right)\)
\(\text{Thay }\widehat{yBC}=140^o\text{ vào }\left(2\right)\)
\(\Rightarrow140^o+\widehat{BCz}=180^o\)
\(\Rightarrow\widehat{BCz}=180^o-140^o\)
\(\Rightarrow\widehat{BCz}=40^o\)
\(\cdot\text{Lại có: }\widehat{ACz}+\widehat{BCz}=\widehat{ACB}\)
\(\text{Mà }\widehat{ACz}=60^o;\widehat{BCz}=40^o\)
\(\Rightarrow60^o+40^o=\widehat{ACB}\)
\(\Rightarrow\widehat{ACB}=100^o\)
\(\text{Vậy }\widehat{ACB}=100^o\)
\(\text{Tổng quát: Nếu }\widehat{A}+\widehat{B}+\widehat{C}=360^o\text{ thì ta luôn chứng minh được Ax//By/Cz}\)
\(\text{Và nếu Ax//By//Cz và biết 2 trong 3 góc thì ta luôn tìm ra được 1 góc còn lại. }\)