Pythagoras of Samos (c570-495BC) was a philosopher and mathematician
and formed a cult relating mathematics to everything. Actually, very
little is known about him. We cannot be sure that he actually
discovered the proof of the relationship between the squares of the
sides of a right angled triangle, or whether it was one of his
successors. Like many labels in mathematics, calling the theorem (or
identity) Pythagoras' Theorem is a convenient way of referring to it.
Pythagoras Theorem states:
In a right angled triangle, the square of the hypotenuse is equal to
the sum of the squares of the other two sides. Two useful right angled
triangles are those with sides, 1, 1, 2, and 3, 4, 5. Of course most
right angled triangles have sides that are not integers, but irrational
numbers.
Proof 1 and Proof 2 below are basically the same, with Proof 2 using more modern labelling.
Proof 1
The sketch on the right can be used as a guide to proving the theorem
geometrically. Another proof is here.
As with many geometric proofs, we need to make the appropriate
construction lines.
Some assumptions
First, ACDE is the square on the hypotenuse, AC. The square BCFG is the
square on the side BC. And the square ABJH is the square of the side AB.
BXY is a line from B at right angles to AC and DE.
BXY is therefore parallel to CD and AE. (Because the sum of the
internal angles of the diagonal DE, and those of the diagonal AC are
180°)
Prove the area of the square on BC is equal to the area of
the rectangle CDYX
We note that the area
of ΔACF is equal to half the area of the square BCFG,
because the area of the square BCFG is BC·CF. And the area of
ΔACF, which has a height, BC, is BC·CF/2.
Similarly, we note that the area of the triangle BCD is equal to half
the area of the rectangle CDYX. This is because the area of
ΔBCD, height CX, is CD·CX/2 and the area of the
square CDYX is CD·CX
Now if ΔACF is equal in area to ΔBCD, then the area
of square CB is equal to the area of rectangle CDYX.
This is indeed the case because ΔACF and ΔBCD each
have sides equal to the length of AC, the hypotenuse and sides equal to
CB. The angle between these is 90° + angle ACB, so the
triangles are congruent (side angle side, SAS). So the area of the
rectangle CDYX is equal to the are of the square on CB.
Prove the area of the square on AB (the other side) is equal
to the area of the rectangle AXYE
We draw the lines JC and BE. As before we note the area of ΔACJ is half
the area of the square on AB, because (as before) the height of the
triangle is equal to the side of the square and the base is the same as
the side of the square.
Similarly,
ΔABE is equal to half the area of AXYE, because the area of ΔABE is
AX·AE/2, which is half the area of the rectangle AXYE.
As
before, if ΔACJ is equal to ΔABE, then the area of AXYE is equal to the
area of the square on AB. Also, as our previous argument, the triangles
have equal sides and equal enclosed angles (both are 90° + angle
BAX).
Conclusion
Because the area CDYX is equal to the square on BC and the area AXYE is
equal to the square on AB, the square on AC is equal to the sum of the
squares on AB and BC ■
Proof 2
This is really a repetition of the previous proof, using different labelling. The sketch is made exactly as before.
So, we represent the sides of the triangle ABC using the small letters, as shown. So the side AC is b.
We wish to show that: a2+c2=b2 [2.01]
Drop a perpendicular from B, through X and Y, cutting AC at X, and CX=b-x and AX=x.
(First, considering the blue lines) The area of ΔACF=a2/2 (Half the base, a, times the height). [2.01]
The area of ΔBCD=b(b-x)/2 (Half the base, b, times the height, b−x). [2.02]
ΔACF and ΔBCD are congruent because, CB=CF (=a), and CD=AC (=b), and angle ACD=angle ACF (=ACB+90°)
So a2/2=b(b-x)/2 a2/2=(b2−bx)/2 Multiplying throughout by 2: a2=(b2−bx) Making bx the subject, we have: bx=(b2−a2) And dividing by b, to get an expression for x: x=(b2−a2)/b [2.02]
(Consider the fuchsia lines) The area of ΔACJ is c2/2
The area of ΔABE is bx/2 Because these triangles are congruent (as before) c2/2 =bx/2 So multiplying throughout by 2, to remove the 2: c2 =bx And making x the subject: x=c2/b [2.03] Equating 2.02 and 2.03, we have: (b2−a2)/b=c2/b
Which rearranged, becomes: (b2−a2)=c2 And making b2 the subject, we have Pythagoras Theorem: b2 =a2+c2 ■
This is really a repetition of the previous proof, using different labelling.