To find the smallest number, we try $A = 1$. The given equation becomes $1BCD = BCD + 1BC \times D$, hence $1000 = 1BC \times D$. This means that $D$ is a divisor of $1000 = 2^3 \times 5^3$ and so $D \in \{1, 2, 4, 5, 8\}$, because $D$ is a digit. On the other hand, $1BC < 200$ implies that $1000 < 200 \times D$, and so $D > 5$. Therefore we need to have $D = 8$ to find a solution. Because $1000/8 = 125$ we now see that there is a solution with $B = 2$ and $C = 5$. Hence the smallest possible value is $1258$.