For each pair of opposite sides of a cyclic quadrilateral take the larger length less the smaller length. Show that the sum of the two resulting differences is at least twice the difference in length of the diagonals.
Solution
We prove the slightly stronger result that the difference between two opposite sides is at least the difference between the diagonals. Suppose the diagonals meet at X. Then AXB, DXC are similar. Suppose AB = kCD with k ≥ 1. Then BE = kCE and AE = kDE. Suppose CE ≥ DE. Then CD + DE > CE, so CD > CE - DE, so (k-1) CD > (k-1)(CE - DE) or AB - CD > BE - CE - AE + DE = BD - AC.
© John Scholes
jscholes@kalva.demon.co.uk
11 May 2002