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Square From Nowhere: What is this about?
A Mathematical Droodle


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


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Explanation

Copyright © 1996-2008 Alexander Bogomolny

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The applet suggests the following construction of a square:

  Let points P and Q lie on a segment AR and satisfy AP = QR. At points P, Q, R erect perpendiculars PD, QB, RC to AR such that PD = PR, QB = QR, and RC = PQ. Then the quadrilateral ABCD is a square.


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


Buy this applet

The proof below assumes that P and Q lie between A and R. This need not necessarily be the case, but if P and Q are without AR, the proof has to be slightly adjusted.

First consider right triangles APD and AQB:

  AP = QR = QB,
PD = PR = AQ.

It follows that APD = AQB. In particular, AD = AB and ADP = BAQ. Therefore, BAD is right.

Further, define T on the extension of RC such that BT||AR and consider BTC.

  BT = QR = AP,

and

 
CT= RC + RT
 = PQ + QB
 = PQ + QR
 = PR
 = PD.

Hence BTC equals the other two. In particular, BC = AB and ABC is also right. This does show that ABCD is a square.

References

  1. N. A. Court, Mathematics in Fun and in Earnest, Mentor Books, 1961

Copyright © 1996-2008 Alexander Bogomolny

28700994Page copy protected against web site content infringement by Copyscape


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