Let the length of AM and perimeter of △ABC be x and L respectively. Circle O is inscribed in △ABC. Hence △AMO ≡ APO (RHS), so AM=AP=x. In the same way △BNO ≡ △BMO, BN=BM=14－x, and △CNO ≡ △CPO, CN=CP=12－x. The primeter of △ABC is L = AM+BC +AC = 14+16+12 = 42. L is also written as follows: L = { ( AM+BM )+( BN+CN )+( CP+AP ) } = { ( AM+AP )+( BN+BM )+( CP+CN ) } = 2( AM+BN+CP ). Plug AM=x, BN=14－x and CP=12－x into the equation above and simplify it. L = 2{ x+(14－x)+(12－x) } = 2( x＋14－x＋12－x ) = 52－2x. Plug L=42 into the equation and find the value of x. 42 = 52－2x, x=5. So, AM=x=5, BN=14－x=9 and CP=12－x=7. The length of tangent segments to circle O are AM=5, BN=9 and CP=7.