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. A useful right angled triangle is that with sides 3, 4, 5. Of course most right angled triangles have sides that are not integers, but irrational numbers. For instance: 1, 1, √2; and √3, 2 1.

Proof 1 and Proof 2 below are basically the same, with Proof 2 using more modern labelling.

As with many geometric proofs, we need to make the appropriate construction lines.

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°)

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 area of the square on CB.

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).

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:

a

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=a

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

a

a

Multiplying throughout by 2:

a

Making bx the subject, we have:

bx=(b

And dividing by b, to get an expression for x:

x=(b

(Consider the fuchsia lines)

The area of ΔACJ is c

The area of ΔABE is bx/2

Because these triangles are congruent (as before)

c

So multiplying throughout by 2, to remove the 2:

c

And making x the subject:

x=c

Equating 2.02 and 2.03, we have:

(b

Which rearranged, becomes:

(b

And making b

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