For our second fallacious proof, there is:

Every Point Inside a Circle is On the Circle

Consider circle O with point P inside the circle. Choose point R on ray OP so that (OP)(OR) = r squared (r^2), where r is the length of the radius of circle O. Let the perpendicular bisector of PR intersect circle O at points S and T; let M be the midpoint of PR.

See figure:

(I) OP = OM - MP

(II) OR = OM + MR = OM + MP

By multiplying (I) and (II):

(III) (OP)(OR) = (OM - MP)(OM + MP)

By the Pythagorean theorem:

(OM)^2 + (MS)^2 = (OS)^2

or:

(IV) (OM)^2 = (OS)^2 - (MS)^2

Also:

(MP)^2 + (MS)^2 = (PS)^2

or:

(V) (MP)^2 = (PS)^2 - (MS)^2

Now substitute (IV) and (V) into (III) to get:

(OP)(OR) = [(OS)^2 - (MS)^2] - [(PS)^2 - (MS)^2

(VI) (OP)(OR) = (OS)^2 - (PS)^2

Because segment OS is the radius of circle O:

(VII) (OS)^2 = r^2 = (OP)(OR)

Now substitute (VII) into (VI):

(OP)(OR) = (OP)(OR) - (PS)^2

Therefore PS = 0, which implies that point P must be on the circle!

To discover the fallacy in this 'proof', we let OP = a. Therrefore OR = r^2/a. Because r > a and the square of a real number is positive, (r - a)^2 > 0. This can be written as r^2 - 2ra + a^2 > 2ra. Multiplying both sides of this inequality by 1/2a, we get 1/2(r^2/a + a) > r, which is /2 (OR + OP) > r, or OM > r. This imples that point M must be OUTSIDE the circle and that points S and T do not exist. This destroys the fallacious proof.

When this author tried to construct the circle according to the instructions, it was found to be impossible.

Go forward to fallacious proof 3

Go back to fallacious proof 1