"False proof" regarding measures of angles?

Question

I'm trying to recall a "false proof" that I saw once a long time ago. First, what I mean by "false proof" is a seemingly logical set of statements that leads to an obviously false conclusion. The most famous example would be the proof that 2 = 1.

Years ago, I watched a professor develop a proof in class which resulted in "proving" that all angles were 60 degrees! But I can't remember the steps at all, not to mention where the fallacy lies. Has anyone seen this construction before?

Answer 1

There is a false proof that every triangle is isosceles. A diagram accompanies this proof which puts the intersection of two of the construction lines in it inside the triangle when, in fact, this intersection is always on or outside the triangle, and this leads to the false conclusion that any two of the sides of the triangle must always be the same length. Now if this "proof" is carried out twice for two different pairs of sides, then it "follows" that all three sides are equal and thus every triangle is equilateral, so all three angles are always 60 degrees.

Start with any triangle ABC. Construct the perpendicular bisector of AC, with D being the midpoint of AC, and also construct the bisector of angle ABC, and let these two constructed lines intersect at E. (Fudge your drawing so that E lies inside the triangle.) Let F lie on AB so that EF is perpendicular to AB. Let G lie on BC so that EG is perpendicular to BC.
Show triangle FBE is congruent to triangle GBE, and so BF = BG and FE = GE.
Show triangle AED is congruent to triangle CED, and so AE = CE.
Now show triangle FAE is congruent to triangle GCE and so FA = GC.
BF = BG and FA = GC. Add these to get BA = BF + FA = BG + GC = BC. Thus BA = BC in any triangle ABC; i.e. every triangle is isosceles.

(In fact, if ABC is not isosceles, E lies outside of ABC, and only one F and G will lie on a side between two vertices. That is, if F is between A and B, then C will be between B and G. If G is between B and C, then A will be between B and F.)

Answer 2

it somewhat is an isosceles triangle. if C is the vertex, then A = B. A = 3x-6, B = 3x - 6, C = 2x and A + B + C = a hundred and eighty so... (3x - 6) + (3x - 6) + 2x = a hundred and eighty 8x -12 = a hundred and eighty 8x = 192 x = 24 so B = (3*24) -6 = sixty six attempt by utilising sixty six+sixty six+40 8 = a hundred and eighty

Answer 3

A+B+C=180

so take A=B=60

C=60 follows

but 3 unknowns in 1 equation, valid in certain cases. But False!