Download The Pythagorean Theorem Figure 1: Given a right triangle ABC with

The Pythagorean Theorem
B
c
a
C
A
b
Figure 1:
Given a right triangle △ABC with vertices A, B, C opposite sides of length
a, b, c, respectively, and 6 C = 90◦ a right angle, the Pythagorean Theorem says
that
c2 = a2 + b2 .
Here is a simple proof.
Proof of the Pythagorean Theorem. Without loss of generality we may
assume that the sides are labeled so that a ≤ b.
c2 is the area of a square with sides c, so let us begin by drawing a square
with sides c. Draw a line through each corner of the square as in fig. (2.i)
below, so that the angle between the line and the corresponding side of the
square is congruent to 6 A. Now 6 C = 90◦ and 180◦ = 6 A + 6 B + 6 C so
90◦ = 6 A + 6 B. Therefore the lines divide each corner of the square into two
angles, one congruent to 6 A and the other congruent to 6 B. (See fig. (2.ii)).
c
A
A
c
A
a
c
B
b
c
b−a
B
c
A
B
b−a
b−a
a
A
A
a
b−a
b
b
c
b
A
a
c
B
c
i)
A
ii)
Figure 2:
It follows that each of the four triangles in in fig. (2.ii) has a side of length
c joining an angle congruent to 6 A with an angle congruent to 6 B. By the
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“angle-side-angle” theorem in elementary geometry it follows that the four triangles in fig. (2.ii) are all congruent to △ABC. Therefore the angles of the
quadrilateral in the center of fig. (2.ii) are all right angles. Clearly the sides of
the quadrilateral all have length b − a so the quadrilateral is a small square with
area (b − a)2 .
Again referring to fig. (2.ii), we see that
(area of big square) = 4(area of triangle) + (area of small square)
hence
c2 = 4
ab
+ (b − a)2 .
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Multiplying out we have
c2 = 2ab + (b2 − 2ab + a2 )
hence
c2 = b2 + a2 .
This completes the proof.
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