a)
Ta có: $\widehat{BAC}+\widehat{MAC}=180^{0}$
$⇒ \widehat{MAC}=180^{0}-\widehat{BAC}=180^{0}-120^{0}=60^{0}$
AD là tia phân gác của $\widehat{BAC}$
$⇒ \widehat{DAC}=\frac{1}{2}\widehat{BAC}=\frac{1}{2}.120^{0}=60^{0}$
$⇒ \widehat{DAC}=\widehat{MAC}=60^{0}$
⇒ AC là tia phân giác của $\widehat{DAM}$
b)
$AD║MC⇒\widehat{DAC}=\widehat{ACM}=60^{0}$
$ΔACM$ có:
$\widehat{MAC}+\widehat{ACM}+\widehat{AMC}=180^{0}$
$⇒ \widehat{ACM}=180^{0}-(\widehat{MAC}+\widehat{ACM})=180^{0}-(60^{0}+60^{0})=180^{0}-120^{0}=60^{0}$
$⇒ \widehat{ACM}=\widehat{AMC}=60^{0}$
