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Theorem

Diagram for geometric proof of en:Ptolemy's theorem. Created in w:Inkscape.

  1. Let ABCD be a en:cyclic quadrilateral.
  2. Note that on the en:chord BC, the en:inscribed angles ∠BAC = ∠BDC, and on AB, ∠ADB = ∠ACB.
  3. Construct K on AC such that ∠ABK = ∠CBD;
    1. Note that since ∠ABK + ∠CBK = ∠ABC = ∠CBD + ∠ABD, ∠CBK = ∠ABD.
  4. Now, by common angles △ABK is en:similar to △DBC, and likewise △ABD ∼ △KBC.
  5. Thus AK/AB = CD/BD, and CK/BC = DA/BD;
    1. Thus AK·BD = AB·CD, and CK·BD = BC·DA;
    2. Adding, (AK+CK)·BD = AB·CD + BC·DA;
    3. But AK+CK = AC, so AC·BD = AB·CD + BC·DA; en:Q.E.D.

Summary

Description Diagram of Ptolemy's theorem
Source Own work
Author User:EnEdC

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Date/TimeThumbnailDimensionsUserComment
current05:49, 28 August 2006Thumbnail for version as of 05:49, 28 August 2006500 × 164 (25 KB)EnEdCDiagram of Ptolemy's theorem

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