Ta có: \(\widehat{BAa}+\widehat{ABb}=180^0\)(trong cùng phía)
\(\Rightarrow\dfrac{1}{2}\left(\widehat{BAa}+\widehat{ABb}\right)=\dfrac{1}{2}.180^0\)
\(\Rightarrow\dfrac{1}{2}\widehat{BAa}+\dfrac{1}{2}\widehat{ABb}=90^0\)
\(\Rightarrow\widehat{A_2}+\widehat{B_1}=90^0\)(do \(\widehat{A_1}=\widehat{A_2}=\dfrac{1}{2}\widehat{BAa},\widehat{B_1}=\widehat{B_2}=\dfrac{1}{2}\widehat{ABb}\))
Ta có: \(\widehat{A_2}+\widehat{B_1}+\widehat{C}=180^0\)(tổng 3 góc trong tam giác)
\(\Rightarrow\widehat{C}=180^0-\widehat{A_2}-\widehat{B_1}=180^0-90^0=90^0\)
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