Answer
AM = 5
BN = 7
PC = 9
Work Step by Step
We use the properties of inscribed circles as well as the segment addition postulate:
$BM + AP = 12 $
$ CN + AP = 16$
$ CN + BM = 14$
Thus, we obtain:
$ 16-CN + 14 - CN = 12 \\ CN = 9 $
Thus, it follows;
$ AP = 12 - 9 = 3$
$BM = 12 - 3 = 9$
Thus, we find the missing sides:
$BN = 16 - 9 = 7 \\AM = 14 - 9 = 5 \\ PC = 12 - 3 = 9$
