4. For the convex quadrilateral $A B C D$, the four vertices satisfy: the product of the distances from each vertex to the other three vertices is equal. Then the quadrilateral $A B C D$ is ( ).
(A) Isosceles trapezoid
(B) Rhombus
(C) Rectangle
(D) Square
4.C.

From the problem, we have
$$
\begin{array}{l}
A B \cdot A C \cdot A D=B A \cdot B C \cdot B D \\
=C A \cdot C B \cdot C D=D A \cdot D B \cdot D C . \\
\text { Then }\left\{\begin{array}{l}
A C \cdot A D=B C \cdot B D, \\
A B \cdot B D=A C \cdot C D, \\
C A \cdot C B=D A \cdot D B .
\end{array}\right.
\end{array}
$$
(1) $\times$ (3) gives $A C^{2}=B D^{2}$.

Since $A C>0, B D>0$, we have
$$
A C=B D \text {. }
$$

Substituting equation (4) into equations (1) and (3) respectively, we get
$$
A D=B C, A B=C D \text {. }
$$

Thus, quadrilateral $A B C D$ is a parallelogram.
Furthermore, from equation (4), we know that $\square A B C D$ is a rectangle.