Question

如图所示,直线a经过正方形ABCD的顶点A,分别过顶点B,D作DE⊥a于点E,BF⊥a于点F,若DE=8,BF=5,则EF的长为

求详解

Answer 1

AFB-DEA相似
BF:AF=EA:DE
5:AF=(EF-AF):8
AF(EF-AF)=40
(EF-AF)^2 +ED^2 = AF^2 + FB^2
(EF-AF)^2 - AF^2 = 25-64 = 39
AF(EF-AF) = 40
EF(EF-2AF) = 39
解：AF=5，EF=13