$18 \cdot 30$ Among the following propositions, which one is true?
(A) A quadrilateral is a parallelogram if one pair of opposite sides are equal and one diagonal bisects the other diagonal.
(B) A quadrilateral is a parallelogram if one pair of opposite sides are equal and both diagonals are equal.
(C) If the quadrilateral formed by connecting the midpoints of the sides of a quadrilateral is a rectangle, then the original quadrilateral is a rhombus.
(D) If the sum of the squares of one pair of opposite sides of a quadrilateral equals the sum of the squares of the other pair of opposite sides, then the diagonals of the quadrilateral are perpendicular.
(Huanggang Region, Hubei Province, China Junior High School Mathematics Competition, 1989)
[Solution] As shown in Figure (1), consider quadrilateral $A B C^{\prime} D$ satisfying the problem's conditions, but it is not a parallelogram. Therefore, option (A) is incorrect.

Consider isosceles trapezoid $A B C D$ (as shown in Figure (2)), then option (B) is incorrect. Consider kite $A B C D$ (as shown in Figure (3)), then option (C) is incorrect.
Examine Figure (4), if $A B^{2}+C D^{2}=A D^{2}+B C^{2}$,
then $A B^{2}-B C^{2}=A D^{2}-C D^{2}$.
Draw $B O \perp A C$ at $O$, then $A B^{2}-B C^{2}=A O^{2}-C O^{2}$.
Draw $D O^{\prime} \perp A C$ at $O^{\prime}$, then $A D^{2}-C D^{2}=A O^{\prime 2}-C O^{\prime 2}$.
From the previous equation, we have $A O^{2}-C O^{2}=A O^{\prime 2}-C O^{\prime 2}$, hence $O^{\prime}$ coincides with $O$. Therefore, option (D) is correct. Hence, the answer is (D).