The diagonals of a convex quadrilateral meet at right angles at X. Show that the four points obtained by reflecting X in each of the sides are cyclic.
Solution
If we shrink the reflected points by 1/2 about the point X, then we get the feet of the perpendiculars from X to the sides. So it is sufficient to show that these four points are cyclic. Let the quadrilateral be ABCD. Let the feet of the perpendiculars from X to AB, BC, CD, DA be P, Q, R, S respectively.
XPBQ is cyclic, so ∠XQP = ∠XBP. Similarly, XPAS is cyclic, so ∠XSP = ∠XAP. But XBP and XAP are two angles in the triangle XAB and the third is ∠AXB = 90o. Hence ∠XQP + ∠XSP = 90o. Similarly ∠XQR + ∠XSR = 90o. Adding, ∠PQR + ∠PSR = 180o, so PQRS is cyclic.
© John Scholes
jscholes@kalva.demon.co.uk
23 Aug 2002