1. For four numbers $a, b, c, d$ it is satisfied: $d>c, a+b=c+d, a+d<b+c$. Arrange these four numbers in order of magnitude.
Solution. From $a+d<b+c$, it follows that $a+d+b<2b+c$. Since $a+b=c+d$, it follows that $c+2d<2b+c$, which means $d<b$.

Similarly, from $a+d<b+c$, it follows that $a+d+c<b+2c$. Since $a+b=c+d$, it follows that $2a+b<b+2c$, which means $a<c$.

Finally, from the condition we have $c<d$ and since $d<b$ and $a<c$, we get $a<c<d<b$.