Question

等腰Rt△ABC的直角边AB=2,点P从A点出发,沿射线AB运动,

Answer 1

DE的长度不改变.

证明:作QH垂直AC的延长线于H,则∠QCH=∠ACB=45°=∠A;

又∠QHC=∠AEP=90°,QC=AP,则:⊿QHC≌ΔAEP(AAS),得QH=AE=PE;CH=AE.

PE垂直AC,QH垂直AC,则QH∥PE,故DH/DE=QH/PE=1,得DH=DE.

即:DC+CH=DC+AE=DE,故DE=AC/2=[√(AB^2+BC^2)]/2=√2.