File:Ptolemy Theorem az.svg - Wikimedia Commons wikimedia.org. Let β be ∠CAD. Ptolemy lived in the city of Alexandria in the Roman province of Egypt under the rule of the Roman Empire, had a Latin name (which several historians have taken to imply he was also a Roman citizen), cited Greek philosophers, and used Babylonian observations and Babylonian lunar theory. (1)\triangle EBC \approx \triangle ABD \Longleftrightarrow \dfrac{CB}{DB} = \dfrac{CE}{AD} \Longleftrightarrow AD\cdot CB = DB\cdot CE. Ptolemy's theorem implies the theorem of Pythagoras. Ptolemy's Theorem | Brilliant Math & Science Wiki cloudfront.net. Consider a circle of radius 1 centred at AAA. In Euclidean geometry, Ptolemys theorem is a relation between the four sides and two diagonals of a cyclic quadrilateral. Ptolemy: Dost thou see that all the red lines have the lengths in whole integers? This gives us another pair of similar triangles: ABIABIABI and DBCDBCDBC   ⟹  AIDC=ABBD  ⟹  AB⋅CD=AI⋅BD\implies \frac{AI}{DC}=\frac{AB}{BD} \implies AB \cdot CD = AI \cdot BD⟹DCAI​=BDAB​⟹AB⋅CD=AI⋅BD. He is most famous for proposing the model of the "Ptolemaic system", where the Earth was considered the center of the universe, and the stars revolve around it. The theorem can be further extended to prove the golden ratio relation between the sides of a pentagon to its diagonal and the Pythagoras' theorem among other things. ∠BAC=∠BDC. ( β + γ) sin. □BC^2 = AB^2 + AC^2. \hspace{1.5cm}. If EEE is the intersection point of both diagonals of ABCDABCDABCD, what is the length of ED,ED,ED, the blue line segment in the diagram? BC &= \frac{B'C'}{AB' \cdot AC'}\\ Few details of Ptolemy's life are known. C'D' + B'C' &\geq B'D', We won't prove Ptolemy’s theorem here. We may then write Ptolemy's Theorem in the following trigonometric form: Applying certain conditions to the subtended angles and it is possible to derive a number of important corollaries using the above as our starting point. \ _\squareBC2=AB2+AC2. Ptolemy's Theorem and Familiar Trigonometric Identities. max⌈BD⌉? A Roman citizen, Ptolemy was ethnically an Egyptian, though Hellenized; like many Hellenized Egyptians at the time, he may have possibly identified as Greek, though he would have been viewed as an Egyptian by the Roman rulers. Similarly the diagonals are equal to the sine of the sum of whichever pairof angles they subtend. The equality occurs when III lies on ACACAC, which means ABCDABCDABCD is inscribable. I will also derive a formula from each corollary that can be used to calc… Ptolemy's theorem - Wikipedia wikimedia.org. ⁡. Let α be ∠BAC. But AD=BC,AB=DC,AC=DBAD= BC, AB = DC, AC = DBAD=BC,AB=DC,AC=DB since ABDCABDCABDC is a rectangle. Ptolemy's Theorem states that, in a cyclic quadrilateral, the product of the diagonals is equal to the sum the products of the opposite sides. In case you cannot get a copy of his book, a proof of the theorem and some of its applications are given here. Let ABDCABDCABDC be a random rectangle inscribed in a circle. In the case of a circle of unit diameter the sides of any cyclic quadrilateral ABCD are numerically equal to the sines of the angles and which they subtend. Ptolemys Theorem - YouTube ytimg.com. Thus, the sine of α is half the chord of ∠BOC, so it equals BC/2, and so BC = 2 sin α. \end{aligned}ABCDADBCACBD​=AB′1​=AC′⋅AD′C′D′​=AD′1​=AB′⋅AC′B′C′​=AC′1​=AB′⋅AD′B′D′​.​, AB⋅CD+AD⋅BC≥BD⋅AC1AB′⋅C′D′AC′⋅AD′+1AD′⋅B′C′AB′⋅AC′≥1AC′⋅B′D′AB′⋅AD′C′D′+B′C′≥B′D′,\begin{aligned} They'll give your presentations a professional, memorable appearance - the kind of sophisticated look … Sine, Cosine, and Ptolemy's Theorem; arctan(1) + arctan(2) + arctan(3) = π; Trigonometry by Watching; arctan(1/2) + arctan(1/3) = arctan(1) Morley's Miracle; Napoleon's Theorem; A Trigonometric Solution to a Difficult Sangaku Problem; Trigonometric Form of Complex Numbers; Derivatives of Sine and Cosine; ΔABC is right iff sin²A + sin²B + sin²C = 2 He did this by first assuming that the motion of planets were a combination of circular motions, that were not centered on Earth and not all the same. ⁡. 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