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Austrian Mathematical Olympiad, National Competition (Final Round, part 2)

Geometry Difficulty 4.9 AIME Prove it Austria

Let $A$ , $B$ , $C$ and $D$ be four different points lying on a common circle in this order. Assume that the line segment $AB$ is the (only) longest side of the inscribed quadrilateral $ABCD$ .
Prove that the inequality
$AB + BD > AC + CD$
holds.
(Karl Czakler)

Solution

Let $S$ denote the common point of the diagonals, and let $a = AB$ and $c = CD$ .
Since $ABCD$ is an inscribed quadrilateral, triangles $ABS$ and $DCS$ are similar. It follows that numbers $r$ and $s$ must exist, such that $AS = sa$ , $BS = ra$ , $DS = sc$ and $CS = rc$ hold. The inequality under consideration can therefore be written in the form
$a + ra + sc > sa + rc + c.$
This is equivalent to
$a(1 + r - s) > c(1 + r - s),$
which is certainly correct, since $a > c$ is given and the triangle inequality in $ABS$ implies $1 + r > s$ .

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