In both cases place the square on the coordinate plane where C=(0,0) D=(1,0) A=(1,1) B = (0,1). It is easy to show E=(1/2,1-√(3)/2) and F=(1+√(3)/2,1/2)
One proof is that the slopes of segments CE and CF are equal.
Slope CE = (1-√(3)/2)/(1/2) = 2-√(3)
Slope CF = (1/2)/(1+√(3)/2) = 1/(2+√(3)) = 2-√(3)
Since these slopes are equal the points are on the same line.
Hence C, E, and F are collinear.
The second proof is that F the a size change image of E centered at C.
Let the magnitude k=2+√(3)
Then k times the x coordinate of E = (2+√(3))*(1/2)
= 1+√(3)/2 = x coordinate of F
And k times the y coordinate of E = (2+√(3))(1-√(3)/2)
= 1/2 = y coordinate of F
A point and its image under a size change are collinear with the center of the size change. Hence C, E, and F are collinear.
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Posted by Jer
on 2013-10-25 09:10:13 |
Two more complicated proofs.
