Numbers... without numbers 2

$A+B+C+D = AD$ means that $A+B+C+D=10A+D$ and so $B+C=9A$

By replacing $B+C=D$ we have $D=9A$ and because $D$ is a single digit then $D\leq 9$ and so $A \leq 1$

And because $A$ is a single digit it's either $0$ or $1$ but $0$ is impossible because if so $ABCD$ wouldn't have sense.

So $A=1$ and $D=9$

$AD * DA = ABCD$ becomes $19 * 91 = 1000 + 100B + 10C + 9$ wich becomes: $10B+C=72$

And so we have a set of two equations for $B$ and $C$ :

$B+C=9$

$10B+C=72$

By resolving it: $B=7$ and $C=2$

And the solution is $1729$