Trigonometry/Proof: Pythagorean Theorem< Trigonometry
This theorem may have more known proofs than any other; the book The Pythagorean Proposition contains 370 proofs.
Proof by SubtractionEdit
This proof uses rearrangement. The figure shows two identical large squares of side .
- The top square contains the square on the hypotenuse plus identical right triangles in its four corners.
- On bottom, the same large square holds the squares on the other two sides plus the same four right triangles, now moved to form two rectangles of sides in the bottom corners.
From both identical large squares, the area of the same four right triangles of sides is subtracted (colored). Subtracting the triangles removes the same (colored) area from the equal-area large squares, so the remaining white areas, and , are equal.
...and that's it!
Euclid's proof is much more complex, and relies on subdividing a figure into pieces and showing that they are congruent pieces. It's a fragment of mathematical history. You do not need to remember this proof. In fact if it's the first time you're reading this book it's quite OK to skip over it and go on to "Exercise: A Puzzle Triangle"
Why is this proof here at all? Partly it's to show that there is more than one way of proving things. Partly it's because Euclid took great care to proceed in small steps each of which he had already proved. In the 'proof by subtraction' we are using facts about areas and how pieces fit together that are true, but that we haven't actually proved.
In outline, here is how the proof in Euclid's Elements proceeds. The large square is divided into a left and right rectangle. A triangle is constructed that has half the area of the left rectangle. Then another triangle is constructed that has half the area of the square on the left-most side. These two triangles are shown to be congruent, proving this square has the same area as the left rectangle. This argument is followed by a similar version for the right rectangle and the remaining square. Putting the two rectangles together to reform the square on the hypotenuse, its area is the same as the sum of the area of the other two squares. The details are next.
Let be the vertices of a right triangle, with a right angle at . Drop a perpendicular from 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.
For the formal proof, we require four elementary lemmata (a step towards proving the full proof):
- If two triangles have two sides of the one equal to two sides of the other, each to each, and the angles included by those sides equal, then the triangles are congruent (side-angle-side).
- The area of a triangle is half the area of any parallelogram on the same base and having the same altitude.
- The area of a rectangle is equal to the product of two adjacent sides.
- The area of a square is equal to the product of two of its sides (follows from 3).
Next, each top square is related to a triangle congruent with another triangle related in turn to one of two rectangles making up the lower square.
The proof is as follows:
- Let be a right-angled triangle with right angle .
- On each of the sides squares are drawn, in that order. The construction of squares requires the immediately preceding theorems in Euclid, and depends upon the parallel postulate.
- From , draw a line parallel to and . It will perpendicularly intersect and at and , respectively.
- Join and , to form the triangles .
- are both right angles; therefore are collinear. Similarly for .
- 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 and BD are equal to FB and BC, respectively, triangle ABD must be congruent to triangle FBC.
- Since A is collinear with K and L, rectangle BDLK must be twice in area to triangle ABD, since it shares a height with BK and a base with BD and a triangle's area is half the product of its base and height.
- 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.
- Adding these two results, AB2 + AC2 = BD × BK + KL × KC.
- Since BD = KL, BD* BK + KL × KC = BD(BK + KC) = BD × BC.
- Therefore AB2 + AC2 = BC2, since CBDE is a square.
...and we're done.
- With thanks to the Wikipedia page The Pythagorean Theorem which provided the initial version of this page. See that page for more notes and references.
- (Loomis 1968)
- See for example Mike May S.J., Pythagorean theorem by shear mapping, Saint Louis University website Java applet
- Jan Gullberg (1997). Mathematics: from the birth of numbers. W. W. Norton & Company. p. 435. ISBN 039304002X. http://books.google.com/books?id=E09fBi9StpQC&pg=PA435.
- Elements 1.47 by Euclid. Retrieved 19 December 2006.
- Euclid's Elements, Book I, Proposition 47: web page version using Java applets from Euclid's Elements by Prof. David E. Joyce, Clark University