I've been working on this problem all night and I don't know what I'm doing wrong! I keep on going in circles, I think I'm either missing something or making it to hard. I'm assuming P can be anywhere...
P is a point inside an equilateral triangle ABC. Perpendiculars PD, PE AD PF are dropped from P onto BC, CA and AB respectively Prove that [PAF]+[PBD]+[PCE]=[PAE]+[PCD]+[PBF]