P为矩形ABCD内任意一点，若PA=3，PB=4，PC=5，则PD长是多少？

过P做两边的垂线，交AB、BC、CD、DA于EFGH
ABCD是矩形，所以PE=BF，PF=BE，PG=CF，DF=AE
AP^2=AE^2+BF^2.....①
BP^2=BE^2+BF^2.....②
CP^2=BE^2+CF^2.....③
DP^2=AE^2+CF^2.....④
①-②+③
AP^2-BP^2+CP^2=AE^2+BF^2-(BE^2+BF^2)+BE^2+CF^2
=AE^2+CF^2=DP^2
所以DP^2=AP^2-BP^2+CP^2=9-16+25=18
DP=3√2

解：设AB//DC，AD//BC，A与C，B与D为对角，过P点作直线EG⊥AB、DC，交AB于E，DC于G；作HF⊥AD、BC，交AD于H，BC于F，则根据勾股定理，得
PA^2=AE^2+BF^2......(1)
PB^2=BE^2+BF^2......(2)
PC^2=BE^2+CF^2......(3)
(1)-(2)+(3)，得
AE^2+CF^2=PA^2-PB^2+PC^2=3^2-4^2+5^2=18
PD^2=AE^2+CF^2=18
PD=3√2
答：PD=3√2