19th Eötvös 1912

ABCD is a quadrilateral with vertices in that order. Prove that AC is perpendicular to BD iff AB² + CD² = BC² + DA².

Solution

Let the diagonals intersect at O and put ∠AOB = x. Then AB² = OA² + OB² - 2·OA·OB cos x, CD² = OC² + OD² - 2·OC·OD cos x. So AB² + CD² = OA² + OB² + OC² + OD² - 2 cos x (OA·OB + OC·OD). Similarly, BC² + DA² = OA² + OB² + OC² + OD² + 2 cos x (OA·OD + OB·OC). Thus if x = 90^o, then cos x = 0 and AB² + CD² = BC² + DA². But if x ≠ 90^o, then cos x ≠ 0 and so AB² + CD² and BC² + DA² are on opposite sides of OA² + OB² + OC² + OD² and hence unequal.

19th Eötvös 1912

© John Scholes
jscholes@kalva.demon.co.uk
29 Oct 2003
Last corrected/updated 29 Oct 03