Cho a//b ,[tex]\widehat{A_{1}} = \widehat{A_{2}}[/tex] ; [tex]\widehat{B_{1}}=\widehat{B_{2}}[/tex]
Tính [tex]\widehat{ACB}[/tex]
[tex]\widehat{A_{1}}=\widehat{A_{2}}\Rightarrow \widehat{A}=\widehat{A_{1}}+\widehat{A_{2}}\Leftrightarrow \widehat{A}=2.\widehat{A_{1}} \\ \widehat{B_{1}}=\widehat{B_{2}}\Rightarrow \widehat{B}=\widehat{B_{1}}+\widehat{B_{2}}\Leftrightarrow
Vì a//b và [tex]\widehat{A}[/tex] với [tex]\widehat{B}[/tex] nằm ở vị trí hai góc trong cùng phía nên:
[tex]\widehat{A}+\widehat{B}=180^o\\2.\widehat{A_{1}}+2.\widehat{B_{1}}=180^o\\2.(\widehat{A_{1}}+\widehat{B_{1}})=180^o \widehat{A_{1}}+\widehat{B_{1}}=\frac{180^o}{2}=90^o[/tex]
Vì [tex]\widehat{A_{1}};\widehat{B_{1}};\widehat{ACB}[/tex] là ba góc trong cùng một tam giác nên
[tex]\widehat{A_{1}}+\widehat{B_{1}}+\widehat{ACB}=180^o \\90^o+\widehat{ACB}=180^o \\\Rightarrow \widehat{ACB}=90^o[/tex]
Vậy [tex]\widehat{ACB}=90^o[/tex][/tex]