jmc

We can factor $ab+bc+cd+da$ as $(a+c)(b+d).$ Then by AM-GM, $$(a+c)(b+d)\le \frac{[(a+c)+(b+d){]}^2}{4}=\frac{10^2}{4}=25.$$Equality occurs when $a=1,$ $b=2,$ $c=4,$ and $d=3,$ so the largest possible value is $\boxed{25}.$