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Common Chord and a Tangent: What is this about?
A Mathematical Droodle

 

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Explanation

Copyright © 1996-2010 Alexander Bogomolny

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The applet may suggest the following statement:

 

Let there be two circles (O) and (Q) -- in notations that show their centers. Assume the circles intersect and AB is their common chord. Let BM be a piece of tangent to (O) inside (Q). If Q lies on (O), then AB = BM.


 

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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Proof

Assume Q is on (O). Join QA, QB, and QM. All three are radii of (Q) and hence are equal. Triangles AQB and BQM are isosceles. In addition, their base angles coincide. Indeed, ABM between tangent and chord AB cuts off arc AQB and equals half the angular measure of the latter. Inscribed ABQ that is subtended by arc AQ which is one half of arc AQB, is ½ABM. It follows that BM is the bisector of angle ABM and ABQ = MBQ. ΔAQB = ΔBQM and AB = BM.

References

  1. R. Nelsen, Proofs Without Words, MAA, 1993, p. 18

Copyright © 1996-2010 Alexander Bogomolny

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