解:设AE交CD于F.
CD∥AB,则∠FCA=∠BAC=∠EAC,AF=CF.设AF=CF=X,则DF=8-X.
∵AD²+DF²=AF²,即4²+(8-X)²=X².
∴X=5.即CF=5,DF=3.
作EH⊥Y轴于H,DF∥HE,则DF/HE=AF/AE,即3/HE=5/8,HE=4.8;
故AH=√(AE²-HE²)=6.4.
所以,E点坐标为(4.8, 6.4).
解:设AE交CD于F.
CD∥AB,则∠FCA=∠BAC=∠EAC,AF=CF.设AF=CF=X,则DF=8-X.
∵AD²+DF²=AF²,即4²+(8-X)²=X².
∴X=5.即CF=5,DF=3.
作EH⊥Y轴于H,DF∥HE,则DF/HE=AF/AE,即3/HE=5/8,HE=4.8;
故AH=√(AE²-HE²)=6.4.
所以,E点坐标为(4.8, 6.4).