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Note that $ab+bc+cd+da=46$ factors as $(a+c)(b+d).$ So, let $r=a+c$ and $s=b+d.$ Then $r+s=17$ and $rs=46,$ so by Vieta's formulas, $r$ and $s$ are the roots of $x^2-17x+46=0.$ Thus, $r$ and $s$ are equal to $$\frac{17\pm \sqrt{105}}{2},$$in some order.

We can let $a=\frac{r}{2}+t,$ $c=\frac{r}{2}-t,$ $b=\frac{s}{2}+u,$ and $d=\frac{s}{2}-u.$ Then $$a^2+b^2+c^2+d^2=\frac{r^2}{2}+2t^2+\frac{s^2}{2}+2u^2\ge \frac{r^2+s^2}{2}=\frac{197}{2}.$$Equality occurs when $a=c=\frac{r}{2}$ and $b=d=\frac{s}{2},$ so the minimum value is $\boxed{\frac{197}{2}}.$