The angle at A is the smallest angle in the triangle ABC. The points B and C divide the circumcircle of the triangle into two arcs. Let U be an interior point of the arc between B and C which does not contain A. The perpendicular bisectors of AB and AC meet the line AU at V and W, respectively. The lines BV and CW meet at T. Show that AU = TB + TC.
Solution
Extend BV to meet the circle again at X, and extend CW to meet the circle again at Y. Then by symmetry (since the perpendicular bisectors pass through the center of the circle) AU = BX and AU = CY. Also arc AX = arc BU, and arc AY = arc UC. Hence arc XY = arc BC and so angle BYC = angle XBY and hence TY = TB. So AU = CY = CT + TY = CT + TB.
© John Scholes
jscholes@kalva.demon.co.uk
22 Oct 1998
Last corrected/updated 21 Aug 03