# What is the flaw in this proof that all triangles are isosceles?

What is the flaw in this "proof" that all triangles are isosceles?

One well-known illustration of the logical fallacies to which Euclid's methods are vulnerable (or at least would be vulnerable if we didn't "cheat" by allowing ourselves to be guided by accurately drawn figures) is the "proof" that all triangles are isosceles. Given an arbitrary triangle ABC, draw the angle bisector of the interior angle at A, and draw the perpendicular bisector of segment BC at D, as shown below:

If the angle bisector at A and the perpendicular bisector of BC are parallel, then ABC is isosceles. On the other hand, if they are not parallel, they intersect at a point, which we call P, and we can draw the perpendiculars from P to AB at E, and to AC at F. Now, the two triangles labeled "alpha" in this figure have equal angles and share a common side, so they are totally equal. Therefore, PE = PF. Also, since D is the midpoint of BC, it's clear that the triangles labeled "gamma" are equal right triangles, and so PB = PC. From this it follows that the triangles labeled "beta" are similar and equal to each other, so we have BE+EA = CF+FA, meaning the triangle ABC is isosceles.

All theorems from Euclidean geometry used in the argument are correct. I know this statement is false but I was wondering if anyone knew what the problem was. I'm very confused about solving this question.

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You read the statement somewhere, it would help us if you indicated where you read it. –  Karl Kronenfeld Sep 4 at 0:13
Euclidean and Non Euclidean Geometries by Marvin Jay Greenberg –  Lil Sep 4 at 0:16
Alf van der Poorten once told a story about a class he taught where students found 27 ways (if I remember correctly) of spelling isosceles. –  Rebecca J. Stones Sep 4 at 0:35

The problem is that the perpendicular and the bisector do not intersect inside the triangle, as the picture erroneously suggests. In fact, this fallacy is a proof by contradiction that for non-isosceles triangles the intersection point $P$ is necessarily outside of the triangle.
Ravi Vakil quipped that "geometry is the art of drawing correct conclusions from incorrect pictures". The reason it works is because most geometric arguments are insensitive to many distortions, for instance the fallacy argument works as long as we move the point $P$ inside the triangle. But there are thresholds that once crossed change the picture qualitatively. This is why many arguments are split into cases. Missing a case produces fallacies, for instance forgetting that the radius of the circumscribed circle can lie outside of a triangle leads to a "proof" that in any triangle all angles are acute, and the radius is shorter than the longest side.
If the drawing is made correctly, $P$ needs to be outside, but not too outside, the triangle. If it is far outside, $E$ is on the extension of $AB$ and $F$ is on the extension of $AC$, two plus signs in the argument change to minus signs, and the argument goes through. If $P$ is not too far outside, $E$ is on the extension of $AB$, $F$ is still between $A$ and $C$ (or the other way around), one plus becomes a minus, and the argument fails.