The area of the rectangle AHGJ is double the area of triangle JAC, and the area of square ACLE is double triangle BAE. The two triangles are congruent by SAS. The same result follows in a similar manner for the other rectangle and square. (Katz, 1993)
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The next three proofs are more easily seen proofs of the Pythagorean Theorem and would be ideal for high school mathematics students. In fact, these are proofs that students could be able to construct themselves at some point.
The first proof begins with a rectangle divided up into three triangles, each of which contains a right angle. This proof can be seen through the use of computer technology, or with something as simple as a 3x5 index card cut up into right triangles.
Figure 4
Figure 5
It can be seen that triangles 2 (in green) and 1 (in red), will completely overlap triangle 3 (in blue). Now, we can give a proof of the Pythagorean Theorem using these same triangles.
Proof:
I. Compare triangles 1 and 3.
Figure 6
Angles E and D, respectively, are the right angles in these triangles. By comparing their similarities, we have
and from Figure 6, BC = AD. So,
By cross-multiplication, we get :
II. Compare triangles 2 and 3:
Figure 7
By comparing the similarities of triangles 2 and 3 we get:
From Figure 4, AB = CD. By substitution,
Cross-multiplication gives:
Finally, by adding equations 1 and 2, we get:
From triangle 3,
AC = AE + EC
so
Figure 8
We have proved the Pythagorean Theorem.
The next proof is another proof of the Pythagorean Theorem that begins with a rectangle. It begins by constructing rectangle CADE with B辣^文-论~文.网http://www.751com.cn A = DA. Next, we construct the angle bisector of <BAD and let it intersect ED at point F. Thus, <BAF is congruent to <DAF, AF = AF, and BA = DA. So, by SAS, triangle BAF = triangle DAF. Since <ADF is a right angle, <ABF is also a right angle.
Figure 9
Next, since m<EBF + m<ABC + m<ABF = 180 degrees and m<ABF = 90 degrees, <EBF and <ABC are complementary. Thus, m<EBF + m<ABC = 90 degrees. We also know that
m<BAC + m<ABC + m<ACB = 180 degrees. Since m<ACB = 90 degrees, m<BAC + m<ABC = 90 degrees. Therefore, m<EBF + m<ABC = m<BAC + m<ABC and m<BAC = m<EBF.
By the AA similarity theorem, triangle EBF is similar to triangle CAB.
Now, let k be the similarity ratio between triangles EBF and CAB.
Figure 10
Thus, triangle EBF has sides with lengths ka, kb, and kc. Since FB = FD, FD = kc. Also, since the opposite sides of a rectangle are congruent, b = ka + kc and c = a + kb. By solving for k, we have