Common Chord and a Tangent: What is this about?
A Mathematical Droodle
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Explanation
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Copyright © 1996-2012 Alexander Bogomolny
The applet may suggest the following statement:
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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.
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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 ½∠ABM. It follows that BM is the bisector of angle ABM and ∠ABQ = ∠MBQ. ΔAQB = ΔBQM and AB = BM.
References
- R. Nelsen, Proofs Without Words, MAA, 1993, p. 18
|Activities|
|Contact|
|Front page|
|Contents|
|Store|
|Geometry|
|Eye opener|
Copyright © 1996-2012 Alexander Bogomolny
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