\(TC:\)
\(\Delta ABCcântạiA\)
\(\Rightarrow\widehat{B}=\widehat{C}=65^0\)
\(\Rightarrow\widehat{BAC}=180^0-65^0\cdot2=50^0\)
\(b.\)
\(TC:\widehat{BAC}+\widehat{DAC}=180^0\)
\(\Rightarrow\widehat{DAC}=180^0-50^0=130^0\)
\(\Rightarrow\widehat{DAM}=\dfrac{1}{2}\widehat{DAC}=\dfrac{130^0}{2}=65^0\)
\(KĐ:\widehat{DAM}=\widehat{ABC}=65^0\)
Mà hai góc ở vị trí đồng vị
\(\Rightarrow AM\) // \(BC\left(đpcm\right)\)