est) that when the chord subtending an arc has been given, that chord too can be found which subtends the rest of the semicircle. Pentagon A more interesting example is the relation between the length a of the side and the (common) length b of the 5 chords in a regular pentagon. By completing the square, the relation yields the golden ratio: Side of decagon If now diameter AF is drawn bisecting DC so that DF and CF are sides c of an inscribed decagon, Ptolemy's Theorem can again be applied – this time to cyclic quadrilateral ADFC with diameter d as one of its diagonals: where is the golden ratio. whence the side of the inscribed decagon is obtained in terms of the circle diameter. Pythagoras's theorem applied to right triangle AFD then yields "b" in terms of the diameter and "a" the side of the pentagon is thereafter calculated as As Copernicus (following Ptolemy) wrote, "The diameter of a circle being given, the sides of the triangle, tetragon, pentagon, hexagon and decagon, which the same circle circumscribes, are also given." Proofs Visual proof The animation here shows a visual demonstration of Ptolemy's theorem, based on Derrick & Herstein (2012). Proof by similarity of triangles Let ABCD be a cyclic quadrilateral. On the chord BC, the inscribed angles ∠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 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. The proof as written is only valid for simple cyclic quadrilaterals. If the quadrilateral is self-crossing then K will be located outside the line segment AC. But in this case, AK−CK = ±AC, giving the expected result. Proof by trigonometric identities Let the inscribed angles subtended by , and be, respectively, , and , and the radius of the circle be , then we have , , , , and , and the original equality to be proved is transformed to from which the factor