Three. (This question is worth $35^{\circ}$ points) In the plane quadrilateral $A B C D$, $A B=1, B C=\frac{\sqrt{10}}{2}, C D=\frac{3 \sqrt{2}}{2}, D A=\sqrt{3}$, $\angle D=75^{\circ}$. Find the minimum value of the sum of the distances from a point inside the quadrilateral $A B C D$ to the midpoints of the four sides.
Three, connect $A C, B D$ intersecting at $O$, let $\angle A O D=\theta, \angle A D O=\alpha$. From $A B^{2}+C D^{2}=A D^{2}+B C^{2}=\frac{11}{2}$, we know $\theta=90^{\circ}$, see the figure.
Thus, $O D=\sqrt{3} \cos \alpha$
$$
=\frac{3 \sqrt{2}}{2} \cos \left(75^{\circ}-\alpha\right) \text {. }
$$

From (1) and $\sin 75^{\circ}=\frac{1}{4}(\sqrt{6}+\sqrt{2})$,
$\cos 75^{\circ}=\frac{1}{4}(\sqrt{6}-\sqrt{2})$,
we can get $\operatorname{tg} \alpha=\frac{\sqrt{3}}{3}$, so $\alpha=30^{\circ}$.
Thus, $O D=\frac{3}{2}, A O=\frac{\sqrt{3}}{2}, O C=\frac{3}{2}$,
$O B=\sqrt{B C^{2}-O C^{2}}=\frac{1}{2}$.
Therefore, $A C=A O+O C=\frac{1}{2}(\sqrt{3})$, $B D=\frac{1}{2}+\frac{3}{2}=2$.

Let the midpoints of $A B, B C, C D, D A$ be $P, Q, R, S$ respectively, and $P R$ intersects $Q S$ at $M$. Clearly, $M$ is the point that minimizes the sum of the distances to $P, Q, R, S$, see the right figure.
Since $A C \perp D B$, 
$P Q R S$ is a rectangle, the minimum value sought is
$$
\begin{array}{c}
M P+M Q+M R+M S \\
=P R+Q S=2 P R \\
=2 \sqrt{P Q^{2}+Q R^{2}} \\
=2 \sqrt{\left(\frac{1}{2} A C\right)^{2}+\left(\frac{1}{2} D B\right)^{2}} \\
=\sqrt{A C^{2}+D B^{2}}=\frac{1}{2} \sqrt{28+6 \sqrt{3}} .
\end{array}
$$
(Provided by Ji Jianbing, Anxin County Teaching Research Office, Hebei Province)