Nice job Henry. Here's a solution without trial and error.
1) from the ones place, we get that B+C=A or 10+A.
2) from the tens place and the fact that the digits are distinct, we get that B+C=10+A. (Alternatively, we could examine each case from step (1), and see that the B+C=A case leads to the 000+000=000 solution. )
3) from (2) and comparing the ones and tens places, we see that B=A+1.
4) substituting A+1 for B gives us A+1+C=A+10, or C=9
5) in the hundreds place we have A+A+1=C, and since C=9, A=4.