已知如图P是长方形ABCD内一点.求证PA²+PC²=PB²+PD²

证明:由P点作AB的平行线交AD于E,交BC于F
∵EF‖AB AB⊥AD
∴EF⊥AD EF⊥BC
设AD=a AB=b AE=m EP=n 则DE=CF=a-m PF=b-n
PA^2=m^2+n^2
PC^2=(a-m)^2+(b-n)^2
∴PA^2+PC^2=m^2+n^2 +(a-m)^2+(b-n)^2
PB^2=m^2+(b-n)^2
PD^2=(a-m)^2+n^2
∴PB^2+PD^2=m^2+n^2 +(a-m)^2+(b-n)^2
∴PA^2+PC^2=PB^2+PD^2
结论成立.本题就是一道勾股定理的运用,你可以体会一下.