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Given a broken chord with as the midpoint of the corresponding arc , then , the foot of the perpendicular from to the longer chord segment , is the midpoint of the broken chord .

Contributed by: Tomas Garza (December 2020)
Open content licensed under CC BY-NC-SA


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Details

The two chords and shown in step 1 may be seen as a line segment "broken" at point . Archimedes proved that if is the midpoint of the arc (step 2), and the foot of the perpendicular from to the longer chord (step 3), then is the midpoint of the broken chord . This is proved by reflecting the green triangle on the segment (step 4) and observing that the resulting triangles and are congruent (the three angles are equal on each of them and they share the side ), so that , and then .

Reference

[1] U. C. Merzbach and C. B. Boyer, A History of Mathematics, 3rd ed., Hoboken, NJ: John Wiley and Sons, 2011 p. 122.



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