Common Chord and a Tangent: 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 https://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| |Eye opener|

Copyright © 1996-2017 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 https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

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 ½∠ABM. 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

|Activities| |Contact| |Front page| |Contents| |Geometry| |Eye opener|

Copyright © 1996-2017 Alexander Bogomolny

 62609025

Search by google: