Question

Top 5 methods of proving Pyth Theorem
Give medals if it is helpful!

Answer 1

At this point in the scientific literature recorded 367 proof of this theorem.
Probably the Pythagorean theorem is the only theorem with such an
impressive body of evidence. This diversity can be explained only by the
fundamental value theorem in geometry.

Answer 2

Through similar triangles:
Let ABC has a right-angled triangle with a right angle C. Draw the altitude
from C and denote its base by H. triangle ACH is similar to triangle ABC in
two corners. Similarly, the triangle CBH is similar to ABC. Introducing the
notation: BC=a Ac=b AB=c. ⇒ a/c=HB/a ; b/c=AH/b ⇒ a²=c*HB ;
b²=c*AH. ⇒ a²+b²=c*(HB+AH)=c² or c²=a²+b².

Answer 3

Proof by equicomplementability

There are four equal right-angled triangles, as shown in the figure.
Quadrilateral with sides c is a square, as the sum of the two acute angles
of 90 °, and the staright angle - 180 °. The total area of the figure is on
the one hand, the area of a square with sides (a + b), and on the other
hand, the sum of areas of four triangles and squares inner square. (a+b)=4*
(ab/2)+c² ⇒ a²+2ab+b²=2ab+c² ⇒ c²=a²+b².

Answer 4

Euclid's proof

Let A, B, C be the vertices of a right triangle, with a right angle at A.
Drop a perpendicular from A to the side opposite the hypotenuse in the
square on the hypotenuse. That line divides the square on the hypotenuse into
two rectangles, each having the same area as one of the two squares on
the legs. Let ACB be a right-angled triangle with right angle CAB. On each
of the sides BC, AB, and CA, squares are drawn, CBDE, BAGF, and ACIH,
in that order. The construction of squares requires the immediately preceding
theorems in Euclid, and depends upon the parallel postulate. From A, draw a
line parallel to BD and CE. It will perpendicularly intersect BC and DE at K
and L, Connect CF and AD, to form the triangles BCF and BDA. Angles
CAB and BAG are both right angles; therefore C, A, and G are collinear.
Similarly for B, A, H + angles CBD and FBA are both right angles;
therefore angle ABD equals angle FBC, since both are the sum of a right
angle and angle ABC. Since AB is equal to FB and BD is equal to BC,
triangle ABD must be congruent to triangle FBC. A-K-L is a straight line,
parallel to BD, then rectangle BDLK has twice the area of triangle ABD
because they share the base BD and have the same altitude BK, Since C is
collinear with A and G, square BAGF must be twice in area to triangle FBC.
Therefore rectangle BDLK must have the same area as square BAGF = AB².
Similarly, it can be shown that rectangle CKLE must have the same area as
square ACIH = AC². Now add results: AB² + AC² = BD*BK + KL*KC. Since
BD = KL, BD*BK + KL*KC = BD(BK + KC) = BD*BC ⇒ AB² + AC² =
BC², since CBDE is a square.
This is a quite long proof and not easy one.

Answer 5

Proof of Leonardo da Vinci
Segment CI divides ABHJ square into two equal parts so
as triangles ABC and JHI equal by construction. Using rotation by 90 degrees
counter-clockwise around the point A, we see equality and CAJI DABG. Area
of painted figures which is equal to the sum of half the area of small
squares built on leg and an area of the original triangle. It is equal to half
square large square built on the hypotenuse plus area of the original triangle.
Thus, half of the amount areas of small squares is equal to half the size of
a large square, and hence the sum of areas of the squares constructed on
the legs is equal to the area of a square constructed on the hypotenuse.

Answer 6

Proof by infinitesimal

da/dc=c/a ; c*dc=a*da ; c dc=a da+b db ; c²=a²+b²+constant
a=b=c=0 ⇒ constant=0. ⇒ c²=a²+b².

Answer 7

@perl

Answer 8

@TheSmartOne @bohotness @sleepyhead314

Answer 9

Good Job :D

Answer 10

*Gave the first medal xD)

Answer 11

ерч

Answer 12

thx

Answer 13

sorry for question marks
i was wrtting on russian language

Answer 14

forgot to smitch languages

Answer 15

Haha, its ok :)