If the sides of a quadrilateral are $a,b,c,d$, prove that the area cannot exceed $(ac+bd)/2$.

Question

MOP 1997: Let $Q$ be a quadrilateral whose side lengths are $a,b,c,d$ in that order. Show that the area of $Q$ does not exceed $(ac+bd)/2$.

My solution: Without loss of generality, let $a$ be the longest side. I first prove that the area cannot exceed (ab+cd)/2. Drawing a diagonal, we can divide the quadrilateral into two triangles with sides are $a$ and $b$, and $c$ and $d$. The triangle with sides $a$ and $b$ can have maximum area if the angle between them is $\pi/2$. In this case, the area between them will be $\frac{1}{2}ab$. Similarly, the area of the other triangle can have a maximum of $\frac{1}{2}cd$. Adding them up, we get that the maximum area of the quadrilateral can be $\frac{1}{2}(ab+cd)$.

Now if this maximum is achieved, which means that the angle between $a$ and $b$, and $c$ and $d$ are both $\pi/2$, by drawing a diagram, we can convince ourselves that $a=c$. This means that $a$ and $b$ are both the longest sides in the quadirlateral. Hence, by the rearrangement inequality, we have $(ab+cd)/2\leq (ac+bd)/2$. Hence, the area of the quadrilateral is less than or equal to $(ac+bd)/2$.

Is the reasoning given above correct? Does there exist a better proof?

Answer 1

Let $d_1$ and $d_2$ be diagonals of the quadrilateral.

Thus, $$S=\frac{1}{2}d_1d_2\sin\measuredangle(d_1,d_2)\leq\frac{1}{2}d_1d_2$$ and by the Ptolemy $$ac+bd\geq d_1d_2\geq2S.$$

Answer 2

Alternatively (and I know that this solution isn't diagrammatic):

As you claimed, the area of a triangle given two sides $a,b$ is maximized when $\gamma=90°$ since you can express the area of a triangle as $$A=\frac{a·b·\sin\gamma}{2}$$ and $\sin\gamma\leq1$ with equality if $\gamma=90°$. Thus the area of the quadrilateral, given the sides $a,b,c,d$ is maximized when this quadrilateral is cyclic. The area $K$ of cyclic quadrilaterals can be expressed as
$$K=\frac{1}{2}·(ac+bd)·\sin\theta\leq \frac{1}{2}·(ac+bd)·1$$