Question

已知：正方形ABCD中,E为BC的中点,F点在DC上且CF=1/4DC,求证：AE⊥EF

Answer 1

连接AF;

设正方形边长为4a;

AB=BC=CD=AD=4a

E为BC的中点;

∴BE=EC=2a;

CF=1/4CD=a;

DF=4a-a=3a;

AE^2=AB^2+BE^2=(4a)^2+(2a)^2=20a^2

EF^2=EC^2+FC^2=a^2+(2a)^2=5a^2

AF^2=AD^2+DF^2=(4a)^2+(3a)^2=25a^2;

AF^2=AE^2+EF^2;

∴AE⊥EF