Editing Ptolemy's theorem (section)

=== Proof by similarity of triangles ===
[[Image:Ptolemy's theorem.svg|thumb|Constructions for a proof of Ptolemy's theorem]]
Let ABCD be a [[cyclic quadrilateral]].
On the [[chord (geometry)|chord]] BC, the [[inscribed angle]]s ∠BAC = ∠BDC, and on AB, ∠ADB = ∠ACB.
Construct K on AC such that ∠ABK = ∠CBD; since ∠ABK + ∠CBK = ∠ABC = ∠CBD + ∠ABD, ∠CBK = ∠ABD.

Now, by common angles △ABK is [[Similarity (geometry)|similar]] to △DBC, and likewise △ABD is similar to △KBC.
Thus AK/AB = CD/BD, and CK/BC = DA/BD;
equivalently, AK⋅BD = AB⋅CD, and CK⋅BD = BC⋅DA.
By adding two equalities we have AK⋅BD + CK⋅BD = AB⋅CD + BC⋅DA, and factorizing this gives (AK+CK)·BD = AB⋅CD + BC⋅DA.
But AK+CK = AC, so AC⋅BD = AB⋅CD + BC⋅DA, [[Q.E.D.]]<ref>{{citation|title=Charming Proofs: A Journey Into Elegant Mathematics|volume=42|series=Dolciani Mathematical Expositions|first1=Claudi|last1=Alsina|first2=Roger B.|last2=Nelsen|publisher=[[Mathematical Association of America]]|year=2010|isbn=9780883853481|page=112|url=https://books.google.com/books?id=mIT5-BN_L0oC&pg=PA112}}.</ref>

The proof as written is only valid for [[simple polygon|simple]] cyclic quadrilaterals. If the quadrilateral is self-crossing then K will be located outside the line segment AC. But in this case, AK&minus;CK = ±AC, giving the expected result.