Question

小弟趣题一道 请各位数学高手接招！！题目如下：

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！！！！！！！！！！！！！！！！~~~~~~~~~~~~~~~~~~~~~~~~~~
请写出详细过程 并标明辅助线 谢谢

Answer 1

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

Answer 2

旋转

Answer 3

几何画板能做出来。