Question

E是正方形ABCD的对角线BD上的一点,EF⊥BC于F,EG⊥CD于G,若连结AE,FG,问AE和FG有何关系?是说明理

Answer 1

AE=FG

用勾股定理证.

连结AC,把AC与BD的交点记为O.

设边长BC=M,BF=N.

则有 EG=FC=M-N,EF=BF=N

BO=（√2/2)M,BE=(√2)N,EO=（√2/2)M - (√2)N =（√2/2)*(M-2N)

AO=（√2/2)M

因为AO⊥BD,FE⊥EG,所以有

AE^2 = AO^2 + EO^2 = (1/2)M^2 + (1/2)(M-2N)^2 = M^2 -2MN + 2N^2

FG^2 = EF^2 + EG^2 = N^2 + (M-N)^2 = M^2 -2MN + 2N^2

所以 AE^2 = FG^2

所以 AE = FG