正方形ABCD边长为5,P为正方形ABCD形内一点,且PC=5,PA=根号5,则PB=? 要有过程

作PE⊥AB,PF⊥BC,垂足分别为E、F,
设PE=m,PF=n,
PA=√5,
根据勾股定理,
PE^2+AE^2=PA^2=5
m+2+n^2=5,(1)
PC=5,
PF^2+CF^2=PC^2,
(5-m)^2+(5-n)^2=25,
25-10m+m^2+25-10n+n^2=25,
m^2+n^2-10(m+n)+25=0,(2)
(1)代入（2）,
5-10（m+n)+25=0,
m+n=3,
m=3-n,(3)
由（3）代入（1）,
(3-n)^2+n^2=5,
9-6n+2n^2=5,
n^2-3n+2=0,
(n-2)(n-1)=0,
n=2,或n=1,
m=1,或m=2,
PB^2=PE^2+PF^2=(5-m)^2+n^2=（5-1）^2+2^2=16+4=20,
PB=2√5,
或PB^2=(5-2)^2+1^2=10,
∴PB=√10,
∴PB=2√5,或PB=√10.