ABC is a triangle. Take points D, E, F on the perpendicular bisectors of BC, CA, AB respectively. Show that the lines through A, B, C perpendicular to EF, FD, DE respectively are concurrent.
Solution
Suppose that the feet of the perpendiculars from A and P to EF are H and K respectively. Then AF2 - AE2 = (AH2 + FH2) - (AH2 + EH2) = FH2 - EH2 = (FH + EH)(FH - EH) = FE(FH - EH). Similarly, PF2 - PE2 = FE(FK - EK). So H and K coincide iff AF2 - AE2 = PF2 - PE2. In other words, P lies on the line through A perpendicular to EF iff PF2 - PE2 = AF2 - AE2.
Thus if P is the intersection of the line through A perpendicular to EF and the line through B perpendicular to FD, then PF2 - PE2 = AF2 - AE2 and PD2 - PF2 = BD2 - BF2. Hence PD2 - PE2 = AF2 - BF2 + BD2 - AE2. But F is equidistant from A and B, so AF2 = BF2. Similarly, BD2 = CD2 and AE2 = CE2. Hence PD2 - PE2 = CD2 - CE2, so P also lies on the perpendicular to DE through C.
© John Scholes
jscholes@kalva.demon.co.uk
11 May 2002