 The third fallacious proof is:

Two segments of unequal length are actually of equal length.

Consider triangle ABC, with MN parallel to BC and MN intersecting AB and AC in poins M and N, respecitvely. We will now prove that BC = MN.

Because MN is parallel to BC, we have triangle AMN ~ triangle

ABC and BC/MN = AB/MN.

It then follows that:

(BC)(AM) = (AB)(MN).

Now, multiply both sides of this equality by (BC - MN) to get:

(BC)^2(AM) - (BC)(AM)(MN) = (AB)(MN)(BC) -

(AB)(MN)^2

By adding (BC)(AM)(MN) - (AB)(MN)(BC) to both sides, we get:

(BC)^2(AM) - (AB)(MN)(BC) = (BC)(AM)(MN) - (AB)(MN)^2

This equation can be written as:

(BC)[(BC)(AM) - (AB)(MN)] = (MN)[(BC)(AM) - (AB)(MN)],

By dividing both sides by the common factor [(BC)(AM) - (AB)(MN)], we find that BC = MN!!

No discussion of mathematical fallacies would be complete without an example of a dilemma from division by zero. We committed this mathematical sin when we divided by zero in the form of [(BC)(AM) - (AB)(MN)], which was a consequence of the triangles proved to be similar earlier.

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