已知任意四边形ABCD,分别以各边作四个正方形,O,P,Q,R分别为四个正方形的对角线交点求证：线段RQ垂直且等于 线段

已知任意四边形ABCD,分别以各边作四个正方形,O,P,Q,R分别为四个正方形的对角线交点求证：线段RQ垂直且等于 线段OP!

这里引用楼上的图.
AG与ID夹角=A1
GB与ID夹角=A1-π/2
BJ与ID夹角=A1-π/2-3π/4+A3+π/4=A1+A2-π
JC与ID夹角=A1+A2-π-π/2=A1+A2+π/2
A=(0,a1)
D=(a1,0)
AG=a2(cos(A1),sin(A1))
G=(a2*cos(A1),a2*sin(A1)+a1)
GB=a2(cos(A1-π/2),sin(A1-π/2))=a2(sin(A1),-cos(A1))
B=(a2*cos(A1)+a2*sin(A1),a2*sin(A1)+a1-a2*cos(A1))
B=(sqrt(2)a2*sin(A1+π/4),a1+sqrt(2)sin(1-π/4))
BJ=a3(cos(A1+A2-π),sin(A1+A2-π))=a3(-cos(A1+A2),-sin(A1+A2))
J=(sqrt(2)a2*sin(A1+π/4)-a3*cos(A1+A2),a1+a2*sqrt(2)sin(A1-π/4)-a3sin(A1+A2))
JC=a3(cos(A1+A2+π/2),sin(A1+A2+π/2))=a3(-sin(A1+A2),cos(A1+A2))
C=(sqrt(2)a2*sin(A1+π/4)-a3*cos(A1+A2)-a3sin(A1+A2),a1+a2*sqrt(2)sin(A1-π/4)-a3sin(A1+A2)
+cos(A1+a2))
C=(sqrt(2)a2*sin(A1+π/4)-a3*sqrt(2)sin(A1+A2+π/4),a1+a2sqrt(2)sin(A1-π/4)-a3*sqrt(2)sin
(A1+A2-π/4))
DC=(sqrt(2)a2*sin(A1+π/4)-a3*sqrt(2)sin(A1+A2+π/4)-a1,a1+a2sqrt(2)sin(A1-π/4)-a3*sqrt(2)
sin(A1+A2-π/4))
DC中点是M=1/2(C+D)
M=1/2(sqrt(2)a2*sin(A1+π/4)-a3*sqrt(2)sin(A1+A2+π/4)+a1,a1+a2sqrt(2)sin(A1-π/4)-a3*sqrt
(2)sin(A1+A2-π/4))
DC=(x,y)  =>  MH=1/2(y,-x)
MH=1/2(a1+a2sqrt(2)sin(A1-π/4)-a3*sqrt(2)sin(A1+A2-π/4),-sqrt(2)a2*sin(A1+π/4)+a3*sqrt
(2)sin(A1+A2+π/4)+a1)) 
H=(sqrt(2)a2*sin(A1)cos(π/4)-a3*sqrt(2)sin(A1+A2)cos(π/4)+a1,-a2*sqrt(2)*sin(π/4)cos(A1)
+a3*sqrt(2)sin(π/4)cos(A1+A2)+a1)
H=(a2*sin(A1)-a3*sin(A1+A2)+a1,a3*cos(A1+A2)-a2*cos(A1)+a1)
IJ=(sqrt(2)a2*sin(A1+π/4)-a3*cos(A1+A2),a1+a2*sqrt(2)sin(A1-π/4)-a3sin(A1+A2))
GH=(a2*sin(A1)-a3*sin(A1+A2)-a2*cos(A1)+a1,a3*cos(A1+A2)-a2*cos(A1)-a2*sin(A1)+a1-a1)
GH=(sqrt(2)*a2*sin(A1-π/4)-a3*sin(A1+A2)+a1,a3*cos(A1+A2)-sqrt(2)*a2*sin(A1+π/4))
因为x(IJ)=-y(GH),y(IJ)=x(GH)
IJ.GH=0
从而IJ⊥GH