\(\Delta\)ABD = \(\Delta\)ACD (c-g-c) \(\Rightarrow\)\(\widehat{A_1}=\widehat{A_2}=10^0\)
lại có \(\widehat{B_1}=\widehat{B_2}=10^0\)(gt)
\(\Rightarrow\Delta ABD=\Delta BAM\left(g-c-g\right)\)
\(\Rightarrow AM=BD\)
bai cho \(\Delta BDCdeu\Rightarrow BD=BC\)
vay BC=AM