Thinking Outside the Triangle 

Given an arbitrary triangle ABC, let D be the foot of the perpendicular from A to BC, let E be the foot of the perpendicular from D to AC, and let F be a point on the line DE, as illustrated below 


Florin Pirvanescu challenged the readers of Mathematics Magazine in June 1991 to prove that AF is perpendicular to BE if and only if FE/FD = BD/DC. Several proofs have appeared, based on synthetic, projective, and vector methods. These proofs are all fairly elaborate, but there is actually a very simple elementary proof, which is a nice example of "thinking outside the box". 

Let G be the foot of the perpendicular from B to DE, as shown below. 


Clearly BDG ~ DAE and DGE ~ DBC, so we have BG/DG = DE/AE and EG/DG = CB/DB. Also, AEF ~ EGB, giving EG/BG = AE/FE, if and only if AF is perpendicular to BE. Thus we have 

_{} 

which implies DE/FE = EG/DG = CB/DB and so FED ~ DBC if and only if AF is perpendicular to BE. 
