4. Cho tam giác ABC có $\widehat{A}$=50$^{\circ}$; $\widehat{B}$=70$^{\circ}$. Tia phân giác góc C cắt AB tại M. Tính $\widehat{AMC}$; $\widehat{BMC}$.
Bài Làm:
Tam giác ABC có: $\widehat{ABC}$ + $\widehat{BAC}$ + $\widehat{ACB}$ = 180$^{\circ}$
Mà $\widehat{BAC}$=50$^{\circ}$; $\widehat{ABC}$=70$^{\circ}$.
Suy ra $\widehat{ACB}$ = 180$^{\circ}$ - $\widehat{ABC}$ - $\widehat{BAC}$ = 180$^{\circ}$ - 50$^{\circ}$ - 70$^{\circ}$ = 60$^{\circ}$
CM là tia phân giác của góc $\widehat{ACB}$ $\Rightarrow $ $\widehat{ACM}$ = $\widehat{BCM}$ = $\frac{1}{2}$.60$^{\circ}$ = 30$^{\circ}$
Xét tam giác AMC có: $\widehat{CAM}$ + $\widehat{AMC}$ + $\widehat{MCA}$ = 180$^{\circ}$
$\Rightarrow $ $\widehat{AMC}$ = 180$^{\circ}$ - $\widehat{CAM}$ - $\widehat{MCA}$ = 180$^{\circ}$ - 50$^{\circ}$ - 30$^{\circ}$ = 100$^{\circ}$
Xét tam giác BMC có: $\widehat{CMB}$ + $\widehat{MBC}$ + $\widehat{BCM}$ = 180$^{\circ}$
$\Rightarrow $ $\widehat{BMC}$ = 180$^{\circ}$ - $\widehat{MBC}$ - $\widehat{BCM}$ = 180$^{\circ}$ - 70$^{\circ}$ - 30$^{\circ}$ = 80$^{\circ}$
