a circle. It's the second of the proofs given by Thbit ibn Qurra. And similar for the segments equal to B'C.
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Pythagorean Theorem and its many proofs - cut-the-knot
Kristen dejting, kristen nätdejting, dejtingsajt för
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More recently, Bi Quang Tuån came up with a different argument: From the above, Area(BA'D) Area(BB'C) and Area(AA'D) Area(AB'C). It makes the algebraic part of proof #4 completely redundant. Erect squares on sides BC and AC as on the diagram. The proof is virtually self-explanatory and the addition of a few lines shows a way of making it formal. Let CHc and BHb be two altitudes nn ABC. Let rectangles ABZ_1Z_2 and ACY_1Y_2 circumscribe the latter two. Note that the segment common to the two squares has been removed.
The circle centered at B and passing through C, in two ways: first, as the square of the tangent AC and then as the product adal: (x y) x(x 2(y z which also simplifies to y 2xz. Let ABC be a right triangle, with the right angle. On the other hand, AEC has AE and the altitude from C equal to AM, where M is the point of intersection of AB with the line CL parallel. Proof #68 The Pythagorean theorem is a direct consequence of the Parallelogram Law. 438-441.) On one hand, the area of the trapezoid equals (2a 2b 2(a b) and on the other, 2ab/2 2ba/2 2c/2. Both a based on the following lemma, which appears to generalize the Pythagorean theorem: Form squares on the sides of the orthodiagonal quadrilateral.