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Euclid I.43 Extended: 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 illustrates an extension of Euclid I.43 that is essentially due to Quang Tuan Bui.

 

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

Through point X inside the parallelogram ABCD draw lines MP||BC and NQ||AB. Euclid I.43 asserts that, if X lies on the diagonal BD then the areas of the off-diagonal parallelograms XNCP and AMXQ are equal. The extension concerns the case where X does not lie on the diagonal. There are two ways to handle this situation. In the applet we distinguish between the cases where X is inside ΔABD and where X is inside ΔBCD. The distinction is made purely for convenience. Both extensions could be demonstrated regardless of the location of X.

Thus, first assuming that X is inside ΔABD, below the diagonal BD. Then Area(XNCP) > Area(AMXQ). The applet suggests a construction that makes up for the difference:

  Let XBYD be a parallelogram, DY intersect NQ in R and RS||BC, with S on AB. Then Area(ASRQ) = Area(XNCP).

It follows that parallelogram MSRX makes up for the difference in areas of XNCP and AMXQ:

 

When X is in ΔBCD, the claim is that again Area(ASRQ) = Area(XNCP), with exactly the same construction.

 

In both case, checking the box "areas on the same side of BD", shows to orange parallelograms. The two also have equal areas. This means (and follows from the fact) that the point of intersection of RS and UV (not shown) lies on the line AX, the diagonal of parallelogram AMXQ. This is just a direct application of Euclid I.43.

 

For a proof (Quang Tuan Bui), let W be the intersection of DR with AB, K is the intersection of RS and CD, E and F the intersections of a line through W parallel to BC with NQ and CD, respectively.

 

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

Then, since SWER and BWRX are parallelograms,

  ER = BW - NR = RX - NR = NX

implying that parallelograms REFK and XNCP are congruent. By Euclid I.43 (in parallelogram AWFD),

  Area(ASRQ) = Area(REFK).

So we get

  Area(ASRQ) = Area(XNCP).

Copyright © 1996-2008 Alexander Bogomolny

28685946Page copy protected against web site content infringement by Copyscape


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