Pythagorean Theorem and its various Proofs

Pythagorean Theorem and its various Proofs

1. Various Proofs of Pythagoras Theorem

2. The Pythagoras Theorem In  mathematics , the Pythagorean theorem or Pythagoras' theorem is a relation in  Euclidean geometry  among the three sides of a  right triangle   ( right-angled triangle ). It states that : In any right-angled triangle, the Square of the Hypotenuse of a Right Angled Triangle Is Equal To The Sum of Squares of the Other Two Sides.

3. The twentieth president of the United States gave the following proof to the Pythagoras Theorem. He discovered this proof five years before he became President. He hit upon this proof in 1876 during a mathematics discussion with some of the members of Congress. It was later published in the  New England Journal of Education. Garfield’s Proof of Pythagoras Theorem (Proof I)

4. In the figure shown below, we have taken an arbitrary right triangle with sides of length a and b and hypotenuse of length c and have drawn a second copy of this same triangle (positioned as pictured) and have then drawn an additional segment to form a trapezium. Garfield’s Proof

5. The parallel sides of the trapezium (which are the top and bottom sides in the figure) have lengths a and b. The height of the trapezium (which is the distance from top to bottom in the figure) is a + b. Thus the area of the trapezium is A = ½ (a + b)(a + b) = ½ (a + b)² However, the area of the trapezium is also the sum of the areas of the three triangles that make up the trapezium. Note that the middle triangle is also a right triangle .The area of the trapezium is thus A = ½ ab + ½ ab + ½ cc = ab + ½ c² Garfield’s Proof (contd.)

6. We thus conclude that ½ (a + b)² = ab + ½ c² Multiplying both sides of this equation by 2 gives us (a + b)² = 2ab + c² Expanding the left hand side of the above equation then gives a² + b² + 2ab = 2ab + c² from which we arrive at the conclusion that a² + b² = c² Hence Proved. Garfield’s Proof (contd.)

7. Proof of Pythagoras Theorem (Proof II) A B C D E F y x u a b c c In the figure, ∆ACB is a right angle triangle, with angle ACB = 90 ⁰ with hypotenuse c To prove : a² + b² = c² Construction: Extend AC to D such that AD = AB = c. Draw ED perpendicular to CD with ED = y Draw AE as the angle bisector of angle BAD. Let EB and EA meet at E. Draw EF perpendicular to CF with EF = x. Proof: In ∆EAD and ∆EAB, AD = AB (by construction) Angle EAD = angle EAB (AE bisects angle BAD) EA is common So, by SAS property ∆EAD is congruent to ∆EAB So, angle ADE = angle ABE = 90⁰ (by CPCTE) and ED = EB = y (by CPCTE)

8. Proof of Pythagoras Theorem ( contd ) A B C D E F y x u a b c c Now, angle EBF + angle EBA + angle ABC = 180⁰ i.e. angle EBF + angle ABC = 90⁰ Also, in ∆EFB, angle EBF + angle BEF = 90⁰ So, angle ABC = angle BEF In ∆ACB and ∆BFE, angle ABC = angle BEF angle ACB = angle BFE = 90⁰ So, by AA similarity ∆ACB is similar to ∆BFE Thus, AC/BF = CB/FE = AB/BE i.e. b/u = a/x = c/y This implies u = bx /a = b( b+c )/a --------(1) and y = cx /a = c( b+c )/a --------(2) but y = u+a (as EFCD is a rectangle) -----(3) So, by using (2), c( b+c )/a = u+a Using (1) we get, c( b+c )/a = b( b+c )/a + a which on simplifying gives a² + b² = c². y

9. W e start with four copies of the same triangle. Three of these have been rotated 90°, 180°, and 270°, respectively . Proof of Pythagoras Theorem (III)

10. Each has area  ab /2. Let's put them together without additional rotations so that they form a square with side  c . Proof of Pythagoras Theorem (III) ( contd )

11. The square has a square hole with the side ( a - b ). Summing up its area ( a - b )² and 2 ab , the area of the four triangles (4· ab /2), we get C²=(a-b)²+2ab C²= a²+b² -2ab+2ab C²=a²+b² Hence Proved. Proof of Pythagoras Theorem (III) ( contd )

12. Theorem: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Given: A right-angled triangle with angle A = 90* Pythagoras Theorem Proof (Through Similarity) (IV)

13. To Prove: (Hypotenuse) 2 = (Base) 2 + (Perpendicular) 2 Construction: From A draw AD perpendicular to BC Proof: In triangles ADC and BAC, ( i ) angle ADC = angle BAC [both 90*] (ii) angle C = angle C [common] By AA similarity criterion, Triangle ADC is similar to BAC. Since corresponding sides are proportional in similar triangles, CD/AC = AC/BC AC 2 = CD X BC (a) In triangles ADB and BAC, ( i ) angle BDA = angle BAC [both 90*] (ii) angle B = angle B [common] Pythagoras Theorem Proof (Through Similarity) (IV) (contd.)

14. So, By AA similarity criterion, Triangle ADB is similar to BAC. BD/AB = AB/BC AB 2 = BC X BD (b) Adding (a) and (b), AB 2 + AC 2 = CD X BC + BC X BD AB 2 + AC 2 = BC( CD + BD) AB 2 + AC 2 = BC(BC) AB 2 + AC 2 = BC 2 Hence Proved Pythagoras Theorem Proof (Through Similarity) (IV) (contd.)

15. Proof by rearrangement (V)

16. Proof by rearrangement (V) ( contd )

17. Pythagoras theorem proof (VI) "The square on the hypotenuse of a right triangle is equal to the sum of the squares on the two legs" (Eves 80-81). This theorem is talking about the area of the squares that are built on each side of the right triangle Accordingly, we obtain the following areas for the squares, where the green and blue squares are on the legs of the right triangle and the red square is on the hypotenuse. area of the green square is area of the blue square is  area of the red square is From our theorem, we have the following relationship: area of green square + area of blue square = area of red square or As I stated earlier, this theorem was named after Pythagoras because he was the first to prove it. He probably used a dissection type of proof similar to the following in proving this theorem. 

18. Architecture and Construction Navigation Earthquake Location Crime Scene Investigation Arrow or Missile Trajectory Some real life uses of Pythagoras Theorem

19. Applications of Pythagoras theorem Pythagoras theorem is used in Coordinate Geometry. It is used in finding the Euclidean distance formula d = (x₂ - x₁)² + (y₂ - y₁)² (x₁, y₁) (x₂, y₂) d = distance a = x₂ - x₁ b= y₂ - y₁ d² = a² + b² = ( x₂ - x₁)² + (y₂ - y₁)² d = (x₂ - x₁)² + (y₂ - y₁)²

20. Pythagorean Triplets (Some observations) One of the Pythagorean triplet is a multiple of 3 One of the Pythagorean triplet is a multiple of 4 One of the Pythagorean triplet is a multiple of 5 Some examples: (3,4,5) (5,12,13) (7,24,25) (8,15,17) (9,40,41) (11,60,61) (12,35,37) (13,84,85) (16,63,65) If you multiply each member of the Pythagorean triplet by n, where n is a positive real number then, the resulting set is another Pythagorean triplet For example, ( 3,4,5) and (6,8,10) are Pythagorean triplets. The only fundamental Pythagorean triangle whose area is twice its perimeter is (9, 40, 41).

21. Applications of Converse of Pythagoras Theorem The converse of Pythagoras theorem can be used to categorize triangles If a² + b² = c² , then triangle ABC is a right angled triangle If a² + b² < c² , then triangle ABC is an obtuse angled triangle If a² + b² > c² , then triangle ABC is an acute angled triangle

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