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 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. <B><U>The Pythagorean Theorem</U></B> <B>The sum of the areas of the squares constructed on the legs of a right triangle is equal to the area of the square constructed on its hypotenuse</B>. <B>Figures 1</B> illustrates the Theorem, showing the right triangle <B>ABC</B>, the squares <B>AGHC</B> and <B>CKMB</B> constructed on its legs and the square <B>ABFE</B> constructed on its hypotenuse. The Theorem states that the area of the square <B>ABFE</B> is equal to the sum of the areas of squares <B>AGHC</B> and <B>CKMB</B>. </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) )}}} <B>Figure 1. A right triangle and squares on its legs and its hypotenuse</B> </TD> </TR> </TABLE> <TABLE> <TR> <TD> <B><U>The Proof of the Theorem</U></B> Let us draw the straight line <B>CL</B> perpendicularly to <B>AB</B> intersecting <B>AB</B> and <B>EF</B> at the points<B>D</B> and <B>L</B> respectively. Then the square <B>ABFE</B> is divided in two rectangles. Let us prove that the rectangle <B>ADLE</B> has the same area as the square <B>AGHC</B>. First, the triangle <B>AEC</B> has the area equal to half of the area of the rectangles <B>ADLE</B>, because the triangle and the rectangle have the common base <B>AE</B>, and the altitude of the triangle <B>AEC</B> to this base is congruent to the rectangle side <B>AD</B>. Second, the triangle <B>AGB</B> has the area equal to half of the area of the square <B>AGHC</B>, because the triangle and the square have the common base <B>AG</B>, and the altitude of the triangle <B>AGB</B> to this base is congruent to the square side <B>AC</B>. Third, the triangles <B>AEC</B> and <B>AGB</B> are congruent having the congruent sides <B>AG</B> and <B>AC</B>, <B>AC</B> and <B>AE</B> (listed by pairs), as well as the congruent angles <B>GAB</B> and <B>CAE</B> 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 <B>CAB</B>. </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) )}}} <B>Figure 2. The proof of the Theorem</B> </TD> </TR> </TABLE> Since the triangles <B>AEC</B> and <B>AGB</B> are congruent, they have same areas. Thus, this chain of arguments proves that the rectangle <B>ADLE</B> has the same area as the square <B>BGHC</B>. Similarly, the rectangle <B>DBFL</B> has the same area as the square <B>CKMB</B>. This implies that the area of the square <B>ABFE</B> is equal to the sum of the areas of the squares <B>AGHC</B> and <B>CKMB</B>. The Theorem is proved. The Pythagorean Theorem has another, equivalent formulation: <B>If <B>a</B> and <B>b</B> denote the lengths of legs of the right triangle and <B>c</B> denotes the length of its hypotenuse, then {{{a^2 + b^2 = c^2}}}</B>. 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 <B>Pythagorean Theorem</B> has many other proofs different from the given in this lesson. Examples of different proofs of the <B>Pythagorean Theorem</B> 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 <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 <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 <A HREF=https://www.algebra.com/algebra/homework/Triangles/GEOMETRY-your-online-textbook.lesson>GEOMETRY - YOUR ONLINE TEXTBOOK</A>.