Pythagorean Theorem and its various Proofs by SamanyouGarg

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									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.
   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.
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.
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
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.
                                                     y
In the figure, ∆ACB is a right angle triangle,   E                   D
with angle ACB = 90 ⁰ with hypotenuse c
To prove: a² + b² = c²                                               c
Construction: Extend AC to D such that           x
AD = AB = c.                                                         A
Draw ED perpendicular to CD with ED = y                      c
                                                                     b
Draw AE as the angle bisector of angle BAD.
Let EB and EA meet at E.                         F   u   B       a   C
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)
Now, angle EBF + angle EBA + angle ABC =                 y
                                                     E                       D
180⁰
i.e. angle EBF + angle ABC = 90⁰
                                                                             c
Also, in ∆EFB,
angle EBF + angle BEF = 90⁰                          x       y               A
So, angle ABC = angle BEF                                            c
                                                                             b
In ∆ACB and ∆BFE,                                    F   u       B       a   C
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.
         Proof of Pythagoras Theorem (III)

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

 The square has a square hole with the side (a -
  b). Summing up its area (a - b)² and 2ab, 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.
       Pythagoras Theorem Proof (Through Similarity) (IV)
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) (contd.)
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
             AC2 = 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.)

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
PROOF BY REARRANGEMENT (V)
PROOF BY REARRANGEMENT (V) (CONTD)
   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.
  Some real life uses of Pythagoras Theorem

Architecture    and Construction
Navigation

Earthquake  Location
Crime Scene Investigation

Arrow or Missile Trajectory
   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)
d² = a² + b²
  = (x₂ - x₁)² + (y₂ - y₁)²                                   d = distance
d = (x₂ - x₁)² + (y₂ - y₁)²          b= y₂ - y₁


                                                                       (x₂, y₂)
                                                     a = x₂ - x₁
 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).
    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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