Is my refutation to this proof in geometry correct?

[neblive]

Here's a proof that all triangles are isosceles (you can ignore my comments to the left, I was too lazy to remove them from the picture)



Of course this proof must not be true, my task is to find the place where this proof falls apart, after lotta tries, I think i've finally found out what's wrong, take a look at this image:


(Sorry the quality sucks cause i'm still a beginner in GeoGebra)
In this image, I've drawn a right angled triangle with sides 3-4-5, I've followed what the proof says
The proof says that AF = AE and FC = BE, this is case 4 cause they meet outside the triangle so
AF - FC = AE - BE
AF - FC = AC
AE - BE is not equal to AB
That's where the proof falls apart
The proof tried to use an unproven statement that says that:

"In case 4, AF - FC = AC and AE - BE = AB"
and this may not necessarily be the case

Did I nail it? or did I make a mistake?

"Due to symmetry, I think this also happens in case 2", is this statement correct?

"As for case 3, I think it is always isosceles", is this statement correct?

Actually cases 2 and 3 can't occur

[Titus]

Error: Graphics such.

Error 2: Is this your homework?

[neblive]

It is not, i'm part-time studying a textbook about euclidean geometry

[DeathRowKitty]

You're correct that case 3 can't occur and case 2 can only occur if the triangle actually is isosceles. You've found an example in which the reasoning they give doesn't work in case 4, but you should ideally be looking to show that the reasoning

always

breaks down in case 4. Otherwise, all you've shown is that the proof can be used to show that at most some-but-not-all non-isosceles triangles are actually isosceles.

[McMenno]

man why do english people have to use all these fancy latin words for math stuff

[neblive]

You're correct that case 3 can't occur and case 2 can only occur if the triangle actually is isosceles. You've found an example in which the reasoning they give doesn't work in case 4, but you should ideally be looking to show that the reasoning

always

breaks down in case 4. Otherwise, all you've shown is that the proof can be used to show that at most some-but-not-all non-isosceles triangles are actually isosceles.

Case 2 can't occur if the triangle is isosceles because it assumes that the angle bisector and the perpendicular bisector aren't coincident.

Since there's a case in which the proof falls apart, then it is an invalid proof.
But i'll find a way to prove that case 4 falls apart completely

[DeathRowKitty]

You're correct that case 3 can't occur and case 2 can only occur if the triangle actually is isosceles. You've found an example in which the reasoning they give doesn't work in case 4, but you should ideally be looking to show that the reasoning

always

breaks down in case 4. Otherwise, all you've shown is that the proof can be used to show that at most some-but-not-all non-isosceles triangles are actually isosceles.

Case 2 can't occur if the triangle is isosceles because it assumes that the angle bisector and the perpendicular bisector aren't coincident.

Since there's a case in which the proof falls apart, then it is an invalid proof.
But i'll find a way to prove that case 4 falls apart completely

Errrr yeah I didn't really pay attention to the line before the description of cases 2 through 4.

tbf it's not a particularly important line but i guess i should really read more closely

You did show that the proof isn't valid by giving a counterexample, but you already knew that every non-isosceles triangle would be a counterexample, so it just seems a bit hollow for what you're doing You did find the part of the proof that breaks down so if you want to consider that as being enough, then that's reasonable, but you can certainly do better

[neblive]

Okay, I think I have proven that if the perpendicular bisector and the angle bisector are not incident then the triangle can't be isosceles (this covers cases 2,3 and 4 even though cases 2 and 3 are impossible):



In the diagram i've drawn an angle bisector through A

If that angle bisector of A is perpendicular to BC, then we can prove that it also bisects BC (l = w = 90 degrees, x = x due to the angle bisector, and we have a common side so both halves of the triangle are congruent, so the the angle bisector bisects the sides as well)

So if that angle bisector is perpendicular to BC then it is coincident with the perpendicular bisector

So we need to prove that if the bisector of angle <A isn't perpendicular then AC =/= AB

In the diagram above, if the angle bisector isn't perpendicular then l =/= w, then since the sum of angles in a triangle equals 180, we find out that z =/= y, therefore AC =/= AB

Q.E.D

[shaft.ed]

It's depressing to remember the things you forgot

[inte]

i refute your refutation

[vonflare]

I refute inte's existence

[inte]

vonflare why are u meming me

[McMenno]

I just realised that your comments are to the right. you liar

[vonflare]

vonflare why are u meming me

because you lost your shitposting touch. You used to be good, but now you're no better than davsto.

[Not_Mafia]

vonflare why are u meming me

because you lost your shitposting touch. You used to be good, but now you're no better than davsto.

Reported

[Annadog40]

But does this mean that squares are just two triangles?

[DeathRowKitty]

Okay, I think I have proven that if the perpendicular bisector and the angle bisector are not incident then the triangle can't be isosceles (this covers cases 2,3 and 4 even though cases 2 and 3 are impossible):



In the diagram i've drawn an angle bisector through A

If that angle bisector of A is perpendicular to BC, then we can prove that it also bisects BC (l = w = 90 degrees, x = x due to the angle bisector, and we have a common side so both halves of the triangle are congruent, so the the angle bisector bisects the sides as well)

So if that angle bisector is perpendicular to BC then it is coincident with the perpendicular bisector

So we need to prove that if the bisector of angle <A isn't perpendicular then AC =/= AB

In the diagram above, if the angle bisector isn't perpendicular then l =/= w, then since the sum of angles in a triangle equals 180, we find out that z =/= y, therefore AC =/= AB

Q.E.D

You seem to have misunderstood what I was saying. I was saying that you should try to prove that that one specific line of the proof with which you found fault is incorrect for all non-isosceles triangles rather than for just one specific such triangle (or that it's wrong some of the time and some other part of the proof is wrong the remainder of the time). Showing that one particular step is wrong for one particular triangle is enough to say that the proof doesn't support their claim, but it's not enough if your goal is to say that that one single thing is the one single thing that causes this proof to sometimes call triangles isosceles when they aren't - it could be that the thing you found is wrong just for that one specific triangle and there could be a different part of the proof that's wrong for literally every other triangle in case 4, for example. (Okay, that really can't be what happens, but you get the idea...)