已知直角坐标系,直线AB分别与X轴,Y轴的正半轴交于点A,B且OA=OB=1.点P（a,b）在第一象限,且2ab=1,由点P向X轴,Y轴所做的垂涎PM,PN（点M,N为垂足）分别与线段AB交于点E,F,试判断三角形AOF与三角形BOE是否一定相似,如果一定相似请加以证明；如果不一定相似或者一定不相似,请简要说明理由~

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OM=a,AM=1-a,AE=√2(1-a)
ON=b,BN=1-b,BF=√2(1-b)
PE=PM-ME=PM-AE=b-(1-a)=a+b-1
PF=PN-NF=PN-BN=a-(1-b)=a+b-1
EF=√2(a+b-1)
BE=BF+EF=√2(1-b)+√2(a+b-1)
=√2(1-b+a+b-1)
=√2a
AF=AE+EF=√2(1-a)+√2(a+b-1)
=√2(1-a+a+b-1)
=√2b
AF/AO=√2b/1=√2*(1/2a)/1=1/√2a=BO/BE
而∠A=∠B=45
所以,三角形AOF与三角形BOE相似