File:Ptolemy's theorem.svg
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Theorem
[edit]Diagram for geometric proof of en:Ptolemy's theorem. Created in w:Inkscape.
- Let ABCD be a en:cyclic quadrilateral.
- Note that on the en:chord BC, the en:inscribed angles ∠BAC = ∠BDC, and on AB, ∠ADB = ∠ACB.
- Construct K on AC such that ∠ABK = ∠CBD;
- Note that since ∠ABK + ∠CBK = ∠ABC = ∠CBD + ∠ABD, ∠CBK = ∠ABD.
- Now, by common angles △ABK is en:similar to △DBC, and likewise △ABD ∼ △KBC.
- Thus AK/AB = CD/BD, and CK/BC = DA/BD;
- Thus AK·BD = AB·CD, and CK·BD = BC·DA;
- Adding, (AK+CK)·BD = AB·CD + BC·DA;
- But AK+CK = AC, so AC·BD = AB·CD + BC·DA; en:Q.E.D.
Summary
[edit]DescriptionPtolemy's theorem.svg | Diagram of Ptolemy's theorem |
Source | Own work |
Author | User:EnEdC |
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current | 05:49, 28 August 2006 | 500 × 164 (25 KB) | EnEdC (talk | contribs) | Diagram of Ptolemy's theorem |
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