**Problem**: (1997 MOP) 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 $ \frac{ac + bd}{2}$.

**Solution**: Let $ A $ be the area of $ Q $. Consider writing $ A $ in terms of $ x,y,z,w $ as shown in the diagram. A common formula for the area of a triangle is equal to half of the product of two sides and the sine of the angle between them. So we have

$ A = \frac{1}{2}(xy\sin{\theta}+wz\sin{\theta}+xw\sin{(180-\theta)}+yz\sin{(180-\theta)}) $.

And since $ \sin{\theta} = \sin{(180-\theta)} $, we get

$ A = \frac{1}{2}\sin{\theta}(xy+wz+xw+yz) = \frac{1}{2}\sin{\theta}(x+z)(y+w) \le \frac{1}{2}(x+z)(y+w) $.

By Ptolemy's Inequality, we know $ ac+bd \ge (x+z)(y+w) $, so then

$ A \le \frac{1}{2}(x+z)(y+w) \le \frac{ac+bd}{2} $

as desired. QED.

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Comment: There should be a geometric, slice-it-up proof to this problem, but the trigonometric proof works out quite nicely as well. Ptolemy's Inequality is pretty useful; the equality case, when the quadrilateral is cyclic, is probably used the most. Whenever a problem has a cyclic quadrilateral with side lengths, Ptolemy is always worth a try. There is a really nice proof using complex numbers (or vectors) and the triangle inequality; see if you can find it.

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Practice Problem: Find the proof of Ptolemy's Inequality mentioned above.

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