$ S_{ABC} = \frac {a \cdot b \sin C} {2} = 1$ ( 1 ) $ CM = a \cdot \cos \frac {C} {2}$ ( 2 ) $ S_{ACM} = \frac {AC \cdot CM \cdot \sin \frac {C} {2}} {2}$ ( 3 ) After substituting (2) in (3), taking into account (1), we get $ S_{ACM} = \frac {1} {2}$. (Here a and b means the lengths of BC and AC respectively), while $ S_{XYZ}$ means the area of $ \Delta XYZ$). Best regards, sunken rock

Extend $ BM$ to meet $ CA$ at $ D$. Then as $ CM$ is both an angle bisector and an altitude, $ DM = MB$. So the perpendicular from $ M$ to $ DC$ will be half the perpendicular from $ B$ to $ DC$, so $ [AMC] = \frac {1}{2}$.