Vì t/g ABC cân tại A (AB=AC)
=> \(\widehat{B}=\widehat{C}\) mà \(\widehat{A}\) = 40*
=> \(\widehat{B}=\widehat{C}\) = (180* - 40*) : 2
=> \(\widehat{B}=\widehat{C}=\) 70*
Xét t/g ABM và t/g ACM có:
BM = CM (M là trung điểm BC)
AB = AC (gt)
\(\widehat{B}=\widehat{C}\left(cmt\right)\)
Do đó: t/g AMB = t/g AMC (c-g-c)
=> \(\widehat{AMB}=\widehat{AMC}\) (2 góc t/ứng)
mà \(\widehat{AMB}+\widehat{AMC}\) = 180* (2 góc kề bù)
=> \(\widehat{AMB}=\widehat{AMC}\) = 180* : 2 = 90*
Xét t/g AMB có:
\(\widehat{B}+\widehat{AMB}+\widehat{BAM}\) = 180*
mà \(\widehat{B}=\) 70* (cmt), \(\widehat{AMB}\) = 90* (cmt)
=> \(\widehat{BAM}=\) 180* - \(\widehat{B}-\widehat{AMB}\)
=> \(\widehat{BAM}=\) 180* - 70* - 90*
=> \(\widehat{BAM}=\) 20*
Vì t/g AMB = t/g AMC (cmt)
=> \(\widehat{AMB}=\widehat{AMC}\) = 90*
và \(\widehat{B}=\widehat{C}\) = 70*
và \(\widehat{BAM}=\widehat{CAM}\) = 20*