all triangles are equilateral!

[phil1882]

if you watch numberphile, then you've probably already seen this "proof" never the less its quite entertaining.

start with three randomly drawn intersecting lines.

label the vertices A,B,C.

take the midpoint of BC and draw a perpendicular line away from A.

we know that two lines from points B and C to this line are of the same length if they meet at the sme point.

draw the angle bisector of BAC.

we know that a line perpendicular to AB that meets this line, is equal to a line that goes from the same  point to AC and is perpendicular to AC.

call the point where the angle bisector, and the perpendicular bisector meet X.

draw lines from B to X and C to X. we know these two lines are equal.

draw a line from X to AB that is perpendicular to AB, call it B* and a line from X to AC that is perpendicular to AC call it C*. these two lines will also be equal.

therefore, By the hypotenuse leg theorem, BXB* = CXC*, and therefore BB* = CC*.

also, by the hypotenuse leg theorem, AXB* = AXC*. and therefore AB* = AC*, and thus AB = AC!

 

your job is to find the flaw. 

 

 

 

[DejMar]

"...draw lines from B to X and C to X. we know these two lines are equal."

This is a flaw. The citation infers that the line BX perpendicular to AB is a right angle to CX where CX is perpendicular to AC for all triangles. While it is true that in Euclidean geometry that, except for the degenerate case where B and C are the same point, a point on the bisector of BAC will be equally distant when measured at right angles from AB and AC, the intersecting points on AB and BC are not necessarily B or C, but can be a different point or points. That is, if we select a point from AB to X at a right angle, the line segment equal in length from AC is not necessarily at point C, except when triangle BAC is an isoceles triangle. In other words, an assumption has been stated for all general triangles when it only applies to isoceles triangles where AB is equal to AC in length.

Edited December 16, 2014 by DejMar

[gavinksong]

The last statement is flawed, BB* = CC* and AB* = AC* doesn't necessarily lead to AB = AC. It relies on the implicit assumption that AB = AB* + BB* and AC = AC* + CC* when it could very well be the case that AB = AB* - BB* or AC = AC* - CC* if B* or C* lies outside of the triangle.

 

This obviously happens whenever AB doesn't actually equal BC, but I haven't thought of a rigorous proof that exactly one of B* or C* falls outside of the triangle when this is the case. However, I do know that the perpendicular bisector of BC and the angle bisector of BAC always intersect outside of the triangle if AB != AC. This can be seen fairly intuitively.

 

Either way, that's the flaw in your reasoning.

[gavinksong]

"...draw lines from B to X and C to X. we know these two lines are equal."

This is a flaw. The citation infers that the line BX perpendicular to AB is a right angle to CX where CX is perpendicular to AC for all triangles. While it is true that in Euclidean geometry that, except for the degenerate case where B and C are the same point, a point on the bisector of BAC will be equally distant when measured at right angles from AB and AC, the intersecting points on AB and BC are not necessarily B or C, but can be a different point or points. That is, if we select a point from AB to X at a right angle, the line segment equal in length from AC is not necessarily at point C, except when triangle BAC is an isoceles triangle. In other words, an assumption has been stated for all general triangles when it only applies to isoceles triangles where AB is equal to AC in length.

 

... or rather, why the step made in the OP is correct is because X also lies on the perpendicular bisector of BC.

[phil1882]

you are indeed correct gavin, that is the incorrect step. the difficult part is proving that's the incorrect step.

[DejMar]

 

A flaw is that when the hypotenuse leg theorem is introduced, it equates the (cross-)product of a point and a line with that of another point and line. In the preparatory statements, A, B, or C are defined as points and B* and C* are defined as lines (i.e., line segments). A point is of zero length, thus 0×B* = 0×C*, but the conclusion BB* = CC* does not follow, as the notation BB* and CC* has not been properly defined. If B* were defined as the other endpoint, along with X of a line segment that could be denoted as XB*, then BB* would then could be assumed to be the notation of a line segment. But then BB* would not be either the hypotenuse or leg of any defined right triangle, and thus the introduction of the hypotenuse leg theorem would not follow. A flaw is then using improper, undefined, or inconsistent notation

[phil1882]

while technically correct, i did not define B* and C* very well, it's not the flaw i am looking for. even if they were defined properly as points, the proof would still stand.

[DejMar]

 

The hypotenuse leg theorem gives that if the hypotenuse of two right triangles are congruent (AX ≡ AX) and a corresponding leg of each are congruent (XB* ≡ XC*), then the triangles are congruent, hence the other legs are also congruent (AB* ≡ AC*). Yet points B and C are not given as equally distant from point A as are the pair of points B* and C*. Furthermore a×b = c×d does not imply |a-b| = |c-d|.

Edited December 22, 2014 by DejMar

[bonanova]

... by the hypotenuse leg theorem, AXB* = AXC* ...

 

Triangles

AXB* and AXC* are not congruent: they are mirror images.

 

The hypotenuse leg theorem requires that corresponding legs be congruent.

The common (congruent) leg AX does not occur in the triangles in a corresponding sense.

[gavinksong]

 

... by the hypotenuse leg theorem, AXB* = AXC* ...

 

Triangles

AXB* and AXC* are not congruent: they are mirror images.

 

The hypotenuse leg theorem requires that corresponding legs be congruent.

The common (congruent) leg AX does not occur in the triangles in a corresponding sense.

 

 

But AX must occur in the corresponding sense, because AX is the hypotenuse of both triangles.

 

I think the flaw was already revealed to be in the last line: "BB* = CC* and AB* = AC*, thus AB = AC".

The mistaken assumption is that BB* and AB* add up to AB, and CC* and AC* add up to AC.

They don't; when a triangle isn't actually iscosceles, only one of the above is an additive relationship and the other is subtractive.

This is because either B* or C* will lie on the triangle, while the other will lie outside of it.

I think Phil is asking us to prove this is always the case.

[phil1882]

gavin has it right bonanova. even presuming axc* and axb* were mirror images, they would sill be congruent however.