7.61 Let $a, b, c$ and $d$ be the lengths of the sides $MN, NP, PQ$ and $QM$ of quadrilateral $MNPQ$, respectively. If $A$ is the area of $MNPQ$, then
(A) $A=\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)$ if and only if $MNPQ$ is a convex quadrilateral.
(B) $A=\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)$ if and only if $MNPQ$ is a rectangle.
(C) $A \leqslant\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)$ if and only if $MNPQ$ is a rectangle.
(D) $A \leqslant\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)$ if and only if $MNPQ$ is a parallelogram.
(E) $A \geqslant\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)$ if and only if $MNPQ$ is a parallelogram.
(28th American High School Mathematics Examination, 1977)
［Solution］Since $A$ is the sum of the areas of the two triangles formed by the diagonal $MP$ of $MNPQ$, that is,
$$
A=\frac{1}{2} a b \sin N+\frac{1}{2} c d \sin Q,
$$

Similarly, $A=\frac{1}{2} a d \sin M+\frac{1}{2} b c \sin P$.
Adding the above two equations, and noting that $0<\sin Q \leqslant 1 \quad\left(0^{\circ}<Q<180^{\circ}\right)$.
Thus, $A \leqslant \frac{1}{4}(a b+c d+a d+b c)=\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)$.
Equality holds if and only if $\sin M=\sin N=\sin P=\sin Q=1$. That is, equality holds if and only if $MNPQ$ is a rectangle.
Therefore, the answer is $(B)$.