Lesson The Pythagorean Theorem

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<H2>The Pythagorean Theorem</H2>
<TABLE> 
 <TR>
 <TD>
The Pythagorean Theorem establishes a remarkable relationship between the sides&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
of any right triangle. This theorem was proved by the Greek geometer Pythagoras
(Pythagoras of Samos), who lived from about 570 B.C. to about 475 B.C., and is 
named after him. This Theorem is the basis and one of corner stones of the 
Euclidean geometry. We give the Theorem and the proof close to as stated by 
Euclid in his "Elements", famous books on Geometry.

The Pythagorean Theorem

The sum of the areas of the squares constructed on the legs of a right triangle&nbsp;&nbsp; 
is equal to the area of the square constructed on its hypotenuse.

Figures 1 illustrates the Theorem, showing the right triangle ABC, 
the squares AGHC and CKMB constructed on its legs and the square ABFE 
constructed on its hypotenuse. The Theorem states that the area of the square 
ABFE is equal to the sum of the areas of squares AGHC and CKMB.
 </TD>
 <TD>
{{{drawing( 330, 345, -3.0, 8.0, -5.5, 6.0, 
 line( 0, 0, 5, 0 ), 
 line( 0, 0, 3.2, 2.4),
 line( 5, 0, 3.2, 2.4),

locate ( 0.1, 0,   A),
            locate ( 5.1, 0,   B),
            locate ( 3.4, 2.5, C),

line( 0,    0,   -2.4, 3.2), 
            line( -2.4, 3.2,  0.8, 5.6), 
            line( 0.8,  5.6,  3.2, 2.4),

locate (-2.7, 3.2, G),
            locate ( 1.0, 5.9, H),

line( 3.2,  2.4,  5.6, 4.2), 
            line( 5.6,  4.2,  7.4, 1.8), 
            line( 7.4,  1.8,  5.0, 0.0),

locate ( 5.8, 4.4, K),
            locate ( 7.4, 1.8, M),

line( 0.0,  0.0,  0.0, -5.0), 
            line( 0.0, -5.0,  5.0, -5.0), 
            line( 5.0,  0.0,  5.0, -5.0),

locate ( 0.0, -5.0, E),
 locate ( 5.0, -5.0, F)
)}}}
Figure 1. A right triangle and squares
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;on its legs and its hypotenuse 
 </TD>
 </TR>
</TABLE> 
<TABLE> 
 <TR>
 <TD>
The Proof of the Theorem

Let us draw the straight line CL perpendicularly to AB intersecting AB and EF at the&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
pointsD and L respectively. Then the square ABFE is divided in two rectangles. Let us 
prove that the rectangle ADLE has the same area as the square AGHC.

First, the triangle AEC has the area equal to half of the area of the rectangles ADLE,
because the triangle and the rectangle have the common base AE, and the altitude of 
the triangle AEC to this base is congruent to the rectangle side AD.

Second, the triangle AGB has the area equal to half of the area of the square AGHC,
because the triangle and the square have the common base AG, and the altitude of 
the triangle AGB to this base is congruent to the square side AC.

Third, the triangles AEC and AGB are congruent having the congruent sides AG and 
AC, AC and AE (listed by pairs), as well as the congruent angles GAB and CAE
concluded between these sides. Congruency of the angles follows from the fact that 
each of the mentioned angles is equal to the right angle plus the common angle CAB.

</TD>
 <TD>
{{{drawing( 330, 345, -3.0, 8.0, -5.5, 6.0, 
 line( 0, 0, 5, 0 ), 
 line( 0, 0, 3.2, 2.4),
 line( 5, 0, 3.2, 2.4),

locate ( 0.1, 0,   A),
            locate ( 5.1, 0,   B),
            locate ( 3.4, 2.5, C),

line( 0,    0,   -2.4, 3.2), 
            line( -2.4, 3.2,  0.8, 5.6), 
            line( 0.8,  5.6,  3.2, 2.4),

locate (-2.7, 3.2, G),
            locate ( 1.0, 5.9, H),

line( 3.2,  2.4,  5.6, 4.2), 
            line( 5.6,  4.2,  7.4, 1.8), 
            line( 7.4,  1.8,  5.0, 0.0),

locate ( 5.8, 4.4, K),
            locate ( 7.4, 1.8, M),

line( 0.0,  0.0,  0.0, -5.0), 
            line( 0.0, -5.0,  5.0, -5.0), 
            line( 5.0,  0.0,  5.0, -5.0),

locate ( 0.0, -5.0, E),
            locate ( 5.0, -5.0, F),

blue(line( 3.2,  2.4, 3.2, -5.0)),
        red(line(-2.4,  3.2, 5.0,  0.0)),
        red(line( 0.0, -5.0, 3.2,  2.4)),

locate ( 3.3, 0.0, D),
 locate ( 3.3, -5.0, L)
)}}}
Figure 2. The proof of the Theorem
 </TD>
 </TR>
</TABLE> 
Since the triangles AEC and AGB are congruent, they have same areas.
Thus, this chain of arguments proves that the rectangle ADLE has the same area as the square BGHC.
Similarly, the rectangle DBFL has the same area as the square CKMB.
This implies that the area of the square ABFE is equal to the sum of the areas of the squares AGHC and CKMB.
The Theorem is proved.

The Pythagorean Theorem has another, equivalent formulation:

If a and b denote the lengths of legs of the right triangle and c denotes the length of its hypotenuse, then {{{a^2 + b^2 = c^2}}}.

This formulation is equivalent to that given at the beginning of this lesson, because {{{a^2}}}, {{{b^2}}} and {{{c^2}}} are areas of the respective squares mentioned in the original formulation.

The Pythagorean Theorem has many other proofs different from the given in this lesson.
Examples of different proofs of the Pythagorean Theorem are presented in the lesson <A HREF = http://www.algebra.com/algebra/homework/Pythagorean-theorem/More-proofs-of-the-Pythagorean-Theorem.lesson> More proofs of the Pythagorean Theorem</A>.

For navigation over the lessons on Properties of Triangles use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/Triangles/Compendium-of-properties-of-triangles.lesson>Properties of Trianles</A>.

For navigation over the lessons on Properties of right-angled triangles use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/Pythagorean-theorem/Properties-of-right-triangles.lesson>PROPERTIES OF RIGHT TRIANGLES</A>.

To navigate over all topics/lessons of the Online Geometry Textbook use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/Triangles/GEOMETRY-your-online-textbook.lesson>GEOMETRY - YOUR ONLINE TEXTBOOK</A>.