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.


What if applet does not run?

Explanation

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 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.


What if applet does not run?

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 ΔAPD = ΔAQB. 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 ΔBTC.

  BT = QR = AP,

and

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

Hence ΔBTC 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

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny

 62020683

Search by google: