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The proof of Thabit’s generalization is identical to one of the standard ways of proving the theorem of Pythagoras using similar triangles, and provides a very accessible generalisation with which to challenge learners.
According to historical documents, it is challenging to establish whether a proof of Th ̄abit’s theorem exists based exclusively on equidecomposibility, as in the case of the Pythagorean and Pappus theorems. This article presents the corresponding proof. Key Words:
Here are the early proofs of the Pythagorean Theorem. Photo by Roman Mager on Unsplash. Ptolemy. While he may be known better for his astronomy, Claudius Ptolemy (b. 85 Egypt d. 165 Alexandria, Egypt) devised one of the first alternate proofs for the Pythagorean Theorem.
By the Pythagorean Theorem (applied twice), the expression |AB |2 + |AC |2 = b 2 + c 2 is equal to 2h 2 + (p + y)2 + (q + y)2. Since the Pythagorean Theorem also yields the equation y 2 + h 2 = r 2, it follows that b 2 + c 2 = 2r 2 – 2y 2 + (p + y) 2 + (q + y) 2 = p 2 + q 2 + 2 p y + 2q y + 2 r 2. Now consider the other side of the equation ...
In the figure, the triangles whose are areas are marked x and y are similar to the original triangle (which has area x+y). So accepting that areas of similar right-angled triangles are proportional to the squares of the hypotenuse, x:y:x+y are in ratio a 2:b 2:c 2, which is Pythagoras's theorem.
the Pythagorean theorem is a special. case, one can write. 52 _|_ C2 = a2 _|_ 26c cos A. 2. Upon substituting for a, b, and c, respectively, in step 1, the result is: (AB)2 + (AC)2 = (BC)2 + 2(AB)(AC) cos A. 3. Since angle BAC = angle AC'B' = angle AB'C, then cos A = cos AB'C = cos AC'B'.
Thabit's Generalization of the Theorem of Pythagoras. Explanation (proof): Can you explain why (prove) Thabit's generalization above is true? Further Exploration: What are the respective conditions under which BC is greater than or smaller than BD + EC?
