All Triangles are Equilateral - Numberphile 

I have an elegant solution, unfortunately this RU-vid comment box is too narrow to contain it.

I love troll math like this. it reminds me of the pi = 4 thing that was going around the internet a few years ago. it's just fun. wrong, but fun.

In fact, all the following propositions are corrects: XB = XC XB* = XC* BB* = CC* XBB* = XCC* AXB* = AXC* AB* = AC* I did it again in a drawing software and I realized that if the line C of the triangle is larger than B, then the point C* is located between the points C and A, and not after the point C, like illustrated. Therefore, in this case the following equations will be: AB* - BB* = AB AC* + CC* = AC Because that, the conclusion "AB=AC" is wrong. The most that we can affirm is: AC = AB + BB* * I'm brazilian and sorry by the poor english.

I FIGURED OUT WHERE YOU WENT WRONG!! ..it's not equilateral

this is learning... an educated man asking amateurs to solve something... I'd love to have my high school classes structured like this

All the congruency is fine since triangles with right angles. Actual mistake starts at 4.05 when it says that AB* - BB* = AB AND AC* - CC* = AC because one of C* and B* is inside and the other one is outside. It can be obviously proven since ABXC is on one circle (X is the middle of the arc BC), so ABX+ACX = 180, hence one of them is bigger that 90 when other is less(they equal only if AB=AC), hence perpendiculars XC* and XB* can't both be inside or both outside,

The problem here is that the example he used here had generalized angles. I tried recreating the problem using a protractor to draw an exact 30-60-90 triangle. I bisected the 90 degree angle and created the same diagram shown above, however, I found that (corresponding) lines on my diagram, XB* and XC*, did not fall on the outside of the triangle! In fact, line XB* fell onto line AB inside the triangle and line XC* fell on line AC outside the triangle! Using the same geometric rules used in the video, I found that line AB (which included the point B*) was equal to the line AC* (which included the point C). Once this happens, we cannot simply subtract the CC* and BB* from both sides, because that would take away part of a side of the triangle ABC! Therefore, this is FALSE!!! :)

Troll level: MATEMATICIAN

Both the points B* and C* cannot lie outside the triangle. The statement AB*=AC* is true, and BB*=CC is true, but for doing AC*-CC* and AB*-BB*, you need to have both the points outside the triangle, which has not been followed here.

I was with him up until 3:21. That's when I started going "No no no no NO NO NO NO NO!"

If you drew this out carefully it'll be really easy to see that if AB is shorter than AC then the point C* will be in between A and C. This means that AB + BB* = AC - CC* and not +. Tricksy

i love this guy's accent

"A proof that all triangles are equilateral - can you see a problem with it?" AB* only equals AC* if point M is on the line AX.

I think the small lie that makes this work is at 3:28. Saying those two triangles are mirror images presumes AB and AC to be equal, which is more or less the thing we are trying to prove here. Its a statement that sneakily begs the question, putting a whole lot of construction lines on there that are equal and hoping you forget the more important parts...

I'll write A° and B° because Google comments use the asterisk to format text into bold text. The mistake is that the picture shows B and C as always being _one the inside_, i.e. the nearest points to A compared to B° and C°. However that's not the case. If B is nearer to A than B°, then C° will be nearer to A than C. Paint a picture in _GeoGebra_ if you want to verify it, but it's what will happen. So everything is correct until the last part, where we don't have AB° = AB + BB° or we don't have AC° = AC + CC°. Which way it is depends on whether AB>AC, or AC>AB (if we knew a priori which one is bigger). To prove this we have to show that if AB>AB° AC AB°. We prove it by contradiction. We assume that AC > AC°. And what we will get (as the video has shown above) is that then AB = AC, which is a contradiction. Since AC != AC° anyway, we have that AC < AC° *Case 2:* AB < AB°. We do the same thing. So we have that *AB > AB° => AC < AC°* and that *AB < AB° => AC > AC°* which is equivalent to the reciprocal of *AB > AB° => AC < AC°*, therefore *AB > AB° AC < AC°*. *PS:* Does someone know how to write asterisks in Google comments without screwing up the text formatting?

I feel that before he started his explanation, he should have shouted "LEEEEEEROOOOOOOOOY NJEEEEEENKINS!!!!

You cannot infer that AXB* ~= AXC*. Sure, B* and C* are 90deg, BB* = CC*, and XB* = XC*, but all we know is (AB* & AC*) > (BB* & CC*). Also, the A angle bisector intercepts MX once. We know MX is an angle bisector of B*XC* because BM = CM and BX =CX. Angle AXC* does not necessarily equal AXB*. AX ix not necessarily an angle bisector of B*XC* because of its one, not infinite, intersection with MX.

The diagram is misleading. In a correct construction, C* would actually be inside AC, so the assumption that |AC| = |AC*| - |CC*| is wrong, and |AC| = |AC*| + |CC*|

I work on automated reasoning in highschool Geometry software, so I know very well where this reasoning chain breaks down. This "paradox" pinpoints exactly what is hidden "under the rug" in Geometry education in highschool: In highschool, we are given an illusion of "rigorous" geometry proof because we are following all the axioms (congruences, angle chasing, etc). But little did we know, these axioms depend on very specific topological facts that does not get proven, but only assumed "experimentally" through inspecting the diagram. So if the diagram is incorrect, the topological assumptions are wrong, and then the axiom applications will result in incorrect facts like so. What topological fact is wrongly assumed in this diagram? It is the fact that both B* and C* are outside of the triangle. In short, what you thought of Geometry in highschool is not rigorous at all, we are doing it semi-formally by mixing axiom applications and experimentally inspecting the diagram (physically interacting with the geometry embedded in the physical universe). To fix this, we have to have axioms that deal with topological statements and make them very explicit when applying the more traditional axioms.

The problem is the positioning of either C* or the B*. If AC was longer than AB, then C* would be within the line AC, so it would be AC*C as apposed to ACC*, and vise verse if AB was longer. He's just very cleverly drawn the diagram very close to an isosceles triangle, so it doesn't look too bad even though it's wrong. If you try and do this with sides that are all very different, you'll see what I mean.

I think the problem lies on assuming that the point X always exists, because if the angle bisector and the perpendicular bisector never meet it can't work surely.

3:36 - Why are the two triangles congruent? Mirror image? It doesn't even look like a mirror image and of course isn't always a mirror image. So we can't say that these two triangles are congruent. If you want to say that it is a mirror image give the axis. But if it is a mirror image (what can happen if AX and MX lie on each other) then the triangle is equilateral. What have I won?

I drew the whole thing in Geogebra to get an accurate image and found out that no matter how I move the original triangle's corners around, the following is the case: Either the point B* lies on |AB| or the point C* lies on |AC|. Unlike in your sketch, they _never both lie outside the original triangle_ (it works in your sketch because you've drawn all angles a little off from what they should be :P). Since you asked us to find the incorrection in your calculations rather than in your picture, here you go: The bottom equation at 4:17 should be AC* + CC* = AC.

I see it this way. You can always draw 2 lines of equal length, that go through some point and two points on some third line, so lines XC and XB can be equal but that doesn't mean that points C and B are at the same distance from A, you can draw another line that goes through X and some point on either of the lines that goes through the AB or AC, that is equal to the XC or XB and at the same distance form the A as either one of them. That said, if you would chose that right line, the rest of the math is correct , but you would work with negative numbers, because the intercession point of the correct line would lie on one of the triangle sides AB or AC, so AB doesn't have to be equal to AC

As many have pointed out, the figure is misleading, C* should be in between A and C. In fact, it can be shown that ABXC is a cyclic quadrilateral, so ABX + ACX = 180. Hence, if ABX is not 90, they will be different, one will be larger than 90 and the other will be less than 90. Resulting either B* in between AB or C* in between AC.

I'm sure someone has worked this out already but there are a few comments denying RHS for congruence and saying that X can't exist so here goes: Given that AC is greater than AB, the point C actually lies further from A than C*. In the above diagram he starts with an almost equilateral triangle and then distorts some angles to make AC shorter than AC*. The proof cannot follow with AC longer than AC*.

I got it: triangle AB*X can't be congruent to triangle AC*X because it doesn't reflect across bisector M properly. MATH!!!

The only flaw is assuming a-b-b* and a-c-c*. Everything else is done correctly, but that means that either AB*-BB*=/=AB or AC*-CC*=/=AC, which destroys the proof.

Ok, so if M is not ON the Angle Bisector of ∠BAC, that means that ∠AXB is not equal to ∠AXC, so if these two are NOT equal, than these two triangles CAN NOT have the same lenght of |AB*| and |AC*|, as |AX| remains the same for both triangles ABX and ACX, and both |C*X| and |B*X| are equal, than all angles have to sum up to 180 degrees, and it can not happen with ∠AXB different from ∠AXC.

Your sketch is imprecise, thus driving us towards assuming that the points are in this order: A - B - B* A - C - C* while actually it's: A - B - B* A - C* - C (or A - B* - B A - C - C*). Assuming that gives us: AB* - BB* = AB AC* - CC* = AC which is incorrect. From the actual point order we get: AB* - BB* = AB AC - CC* = AC* (or AB - BB* = AB* AC* - CC* = AC). Varying sketches due to different arbitrary triangles (their angles, to be specific), whether X exists or not are also different cases we probably should consider, but this is probably what you were going for.

Angle AXB* is not equal to Angle AXC*, so his second congruency--where he says Tri AXB* and Tri AXC* are mirror images--doesn't work. (He also never explains why he assumed they were congruent.) Instead, Angle MXB* is equal to Angle MXC*, since Tri BMX ~= Tri CMX (SSS) and Ang BXB* ~= CXC* (angles in congruent right triangles). If Seg MX coincided with Seg AX then Tri ABC would be isoceles (AB ~= AC). After that, he would have to show that M2Y/M3Z and BY/CZ coincide on another side to prove it equilateral.

I see a lot of over complicated methods of point out the error so I'll pitch in: 1) The perpendicular bisector at M will extend out to point X which is where the bisector angle at A also reaches. This is ok so far. 2) Any angle drawn from the bisector AX to B* or C* must have a common angle and any angle drawn form bisector MX to B or C must have a common angle. 3) Since these lines AX and MX are NOT parallel we know that angle AXB* - MXB != AXC* - MXC. This is because AXB* = AXC* and MXB = MXC but they do not have the same central point. That is to say that *either* part of AXB* overlaps part of MXC or part of AXC* overlaps part of MXB. So in other words, AXB* (union) MXC is greater than or less than AXC* (union) MXB but only equal in the case where AB = AC. 4) The result is that the longer of AB or AC will be the side in which B* or C* is a shortening of that line and the shorter of AB or AC will be an extension. That is to say, that if AB > AC then AB* < AB and AC* > AC. This video depicts both AB* > AB and AC* > AC which is NOT the reality for any triangle that is not isosceles on the side being examined. In the case of a triangle where at least two sides are the same, proving that is impossible since MX will be a subset of AX and thus have infinite points of intersection.

The picture suggests that both B* and C* points will be outside of the triangle. Only then equations written in 4:02 will be correct. The truth is that always one of those points is outside, and the other one is inside (on one of the edges).

The graph they drew is a trick! XBB* is congruent to XCC*. AXB* is congruent to AXC*. However, in their graph the triangle XCC* has the wrong orientation. The point C* should be closer to A than C. So, their equation AC*-CC*=AC is wrong. The correct equation would be AC*+C*C=AC. They were able to manipulate the graph that way because they chose a small difference between AB and AC. When you draw it for yourself with a big difference between AB and AC it should be obvious.

One thing I noticed when messing around with the problem is angle ABX + angle ACX = 180 degrees (or pi radians). I wanted to find a way to mathematically prove that statement because that would have also served as proof that AC* cannot be greater than AC at the same time as AB* is greater than AB. I didn't have much time to spend on it though, so I wasn't able to come up with a proof. I might check on it again later.

This spoof proof can actually be seen through multiple ways, I see as I am reading through the comments. I saw the flaw in this problem when I realized that in a true equilateral triangle, AX and MX should be on top of each other.

AB needs to be equal to AC in order for BB* to be equal with CC*. We know this isn't the case otherwise M would fall on the angle bisector of A. Carlo is assuming the triangle is isosceles in order to prove that the triangle is isosceles. Circular reasoning detected.

Jeez, did anyone actually pay attention in their trigonometry class? You can't declare XBB* and ACC* congruent based on the data we have - the two sides are similar, yes, but it's the angle BETWEEN those similar sides that matters in the proof of congruency, not just any arbitrary angle. (there's a reason why the corresponding rule of congruence is called Side, Angle, Side, not Side, Side, Angle) Picture it like this: imagine that X is a nail that pins two sticks XB and XB* to the wall, and at the end of XB* there's another stick fixed at right angle. With this setup, you can rotate XB* with the protraction at the side and get an infinite number of DIFFERENT triangles that all share those same characteristics. As we can see, this is no basis to proclaim that two triangles are similar. Case dismissed. In the future just try googling basic trigonometry before arguing about the drawings

I know where you went wrong. You were assuming that because the line AX was the angle bisector of Angle A, that triangle AXB* is congruent to AXC*, which is a faulty conclusion. In order for this to be true, Angle AXB* would have to be congruent to Angle AXC*, which would mean that line AX would bisect angle B*XC*, which it does not because line MX bisects that angle by definition of Perpendicular bisector. If a perpendicular bisector of a triangle (in this instance, triangle BXC) hits the opposite point on a triangle, that means it is the angle bisector of that opposite point.

the trick is that the angle at a is nearly a right angle. the two triangles axb* and axc* aren't in fact congruent but are close.

At 4:06, the video states that AB* - BB* = AB and AC* - CC* = AC... But is this true? What if AB > AB* and AC < AC*? The drawing may be throwing us off here... so lets try with some numbers... Lets Say that AB = 4, AC = 7 and BC = 10. That means that BM = CM = 5 According to the cosine rule, a² = b² + c² - 2bc cosA, we know that 10² = 7² + 4² - 2(7*4) cosA cosA = -5/8 A = 128.7° and 4² = 7² + 10² - 2(7*10) cosC cosC = 19/20 C = 18.19° so B = 180°-(128.7°+18.19°) B = 33.11° Lets call the intersection of AX and BC point Q We can solve for the lengths of BN and CN => BQ = 3.6366 CQ = 6.36405 Since BM = CM = 5, we know that M must be between C and Q and we know that QM must be CQ - CM so QM = 1.36405 We also can calculate angle AQC = 97.4597° This means that angle XQC must be 82.5403° therefore, MX = 10.4179 AND QX = 10.5068 We can also get AQ length as 2.20387 => AX length is 12.71067 Next we want to get the lengths of BX and CX. We know that BXM and CXM are right triangles with the MX leg measuring 10.4179 and the BM/CM legs measuring 5, so that means that BX and CX must be 11.5556... This means that the We further know that AB*X and AC*X are right triangles where the hypotenuse is 12.71067 and the XAB*/XAC* angles are 64.35°, so we can use some trigonometry to find that AC* = 12.71067 * cos64.35° = 5.50210... AB* = 12.71067 * cos64.35° = 5.50210... XC* = 12.71067 * sin64.35° = 11.46 XB* = 12.71067 * sin64.35° = 11.46 This lets us find that CC* = sqrt(CX² - XC*²) = sqrt(11.5556² - 11.46²) = 1.48334... BB* = sqrt(BX² - XB*²) = sqrt(11.5556² - 11.46²) = 1.48334... So that would mean that AB* - BB* should equal AB and AC*-CC* should equal AC... right? 5.50210 - 1.48334 ~ 4 So then why would AC = 7 when AC*-CC* is about 4? The answer is that AC* is shorter than AC! So the drawing makes it look like C* should fall farther from A than C, but the reality is that C* does not need to fall farther from A than C, it just needs to be at the nearest point between X and the AC leg of the BAC angle. In the case that AC < AC* and AB < AB*, we would expect our triangle to be isosceles, and in the case that AC > AC* and AB > AB* we would also expect our triangle to be isosceles. And in the case that one of these two conditions is met, and we can also do a similar construct on the B or C vertex where the distance from the vertex to the * point is either greater or less than the other two angles in both cases, then we would expect to find that the triangle is equilateral. But if they are not either both greater or both less, then the proof falls apart on the assumption that AB* - BB* = AB and AC* - CC* = AC

All Triangles are Equilateral - Numberphile AB*X =/= AC*X This is because Line AX is not in the exact center of the triangle.

I’m pretty sure it’s because for a triangle to be equilateral the angles have to be the same but they aren’t the same if your angle bisector and perpendicular bisector don’t line up completely. If the intersect outside the triangle then they not all of the angles are correct and the triangle cannot possibly be equilateral

The error is in the statements at 4:03. To make the statements AB*-BB*=AB and AC*-CC*=AC is to assume that both AB* and AC* are respectively longer than AB and AC, which is not necessarily the case. In fact, unless AB and AC are equal, one of AB* or AC* must be shorter than it's respective AB or AC. The drawing's roughness misleads you into thinking that both B* and C* are outside the original triangle.

I have drawn a (less equal) triangle and everything is correct as presented up to the final calculus conclusion (at 3:55 ). It is stated that AB* - BB* = AB . That is not really the case. My point B* is between A and B (the line AB* is shorter than AB). So the " = AB " part is not always true, I'm not sure why it's wrong in the video, though. But still as suggested, AB* - BB* = AC* - CC* but those arent equal to AB and AC. Thoughts?

With a carefull picture (use compass and ruler) you can easily find out where is the hoax. SPOILER! The bug is in positioning of B* and C*. If the triangle isn't an isosceles triangle (AB AC), exactly one of those points lies between vertices of a triangle ABC. If the triangle is isosceles, B=B* and C=C*. If for instance C* is between A and C, then AC*=AC-CC*, so we don't add, but substract sides. Hence we don't get and paradox etc.

The problem here starts at 3:28 where Carlo says that AXB* is a 'mirror image' of AXC*. However, they are not congruent, because it does not follow the SAS rule. side B*X=C*X (given), side AX = AX (common), but angle B*XA does not equal C*XA. We know this because since MX is the angle bisector of angle BXC, angle BXM must be equal to angle CXM. Therefore, angle B*XM is equal to C*XM (angle BXB* is equal to angle CXC* because of matching angles of congruent triangles). Since B*XM is not the same angle as BXM, and C*XM is not the same angle as CXM, then B*XM cannot be equal to C*XM. Now, since those two angles are not equal, the two green triangles are not congruent or 'mirror images', meaning that AB* is not equal to AC*, and therefore AB is not equal to AC.

Numberphile There's a miss conception there, being AC*=AB* does not mean AB=AC nor does it mean BB*=CC*, it just means that sum AB+BB*=AC+AC*, AB can be different from AC (aswell as BB* be different rom CC*) and the sum can still be the same.So this hipotesis is wrong, you cant say that all triangles have the same length size, you can say however that from any triangle you are able to make an equilateral triangle.

The error is at 2:00 : "Then I know that XB* is equal to XC*" . We do NOT know that! How do we know that? All we know is that XB=XC and that it's a right triangle but we know nothing about how XB* compares to XC*

Everything is fine until 2:45. If you know the length of two sides of a triangle, you need to know the enclosing angle to determine the length of the third side, any other angle won't do. In other words, we cannot say that BB* = CC*, since in the general case BXB* =/= CXC*.

The mistake comes from the approximation of the drawing which puts both B* and C* outside the ABC triangle. They are actually always one inside (closer to A) and one outside the triangle with CC*=BB*=|(AB-AC)/2|.

3:50 -- here is the prpblem. AXB* and AXC* ARE congruent, but though either C or B lies outside AC* or AB* correspondingly, we can't conclude that either AB*-BB*=AB or AC*-CC*=AC. So the main idea is that of of the perpendiculars from x falls inside the triangle. It can be easily prooved just buy drawing approptiate triangle (as an counterexample). In general it's harder to proove it but also possible.

Well for one thing I know, Congruency for side-angle-side is when the angle is include between the two sides... in the video its just showed us the one angle is 90deg, whereas we need to show that angCXC* = angBXB*

Problem is at 2:00 XB* =/= XC* just because they're on the angle bisector. It's easy enough to see with a differently drawn triangle... For example, look at the point where AX and BC intersect: If they were equal, that would mean that the two triangles of triangle ABC have equal heights, which is obviously not true unless B=C. And if you did this for all the sides, it wouldn't be true for them unless A=B and A=C, and if they did, then obviously it's congruent. As some other people pointed out, it only works the way he says it does if AX=MX, which it doesn't.

Thanks for the fun trick question! The poorly drawn diagram shows both AB*>AB and AC*>AC, which is not possible. Instead, if AB*>AB then AC*

Geometry is the art of correct reasoning on incorrect figures -George Polya

*Suspicious Things* Pause at 4:45. That "right angle" in the top left looks mighty obtuse to me! It's clever how the initial "arbitrary triangle" was chosen. The fact it's so close to being isoceles conceals the trickery - just like that illusion where 4 coloured pieces of a "right angled triangle" are shuffled about to open up a mysterious, empty square. It works because the imperfection is imperceivable.

The problem is that AC*X is not actually a mirror image of AB*X. That would only be true if AX bisected the angle B*XC* (and BXC too, of course). Since MX already bisects it, and it's obviously a different line than AX, AC*X and AB*X can't be congruent.

When you bisect an angle in a triangle, that bisecting line will also bisect the opposite side of the triangle, that is, it will cross the opposite side on its middle point (though not perpendicularly). Therefore, point X and point M are always colocated, and from there everything else falls apart.

The problem with this is that when you draw the shortest line between X and AB (AC), B' (C') may be between A and B (A and C). Then the equations in the end are invalid.

You should do a find the bogus step in the astounding -1/12 video. That would be awesome!

In drawing this out myself, I noticed that in my examples C* or B* was within AC or AB, respectively. The fact that you can't have negative values for lines (such as CC* or BB*) should raise a red flag. Additionally, if the angle at A is large enough, this happens with both B* and C* at once. I know this is a problem with the proof, but I can't figure out _why_.

The problem with this (that I found by drawing a few times out on paper) is that either C* or B* will be inside AC or AB. (Say it is C* so I don't need to write everything twice). This means that AC≠AC*-CC* but actually AC=AC*+CC*, while AB=AB*-BB*. If it where the other way around just switch all the Bs and the Cs. It appears that CC*=BB*=AC-AB, making it look as if it works out.

Trick question. There is no mistake in the proof, all triangles are equilateral.

triangles BB*X and CC*X are not congruent. assumption thet they are is false. (you need two pairs of equal sides and the same angle between them to have congruent triangles, any other angle won't do)

I see the flaw. At 3:30, you don't know that angB*XA = angC*XA, thus the two triangles are not necessarily congruent.

The fallacy is at 3:30 , AXC* is in no way a mirror image of AXB* (because AB!=AC, therefore AB*!=AC*)

Every time I tried to draw it I noticed that AC always bigger than AC* , if AC>AB. Same goes for AB if AC

OK I've tried this in a CAD program, and I can tell that the picture they've drawn is misleading. The perpendicular to AC from point X will be to the left of C, not to the right. Because of that, the "proof" fails at assuming congruence between AB*X and AC*X. Because angle AC*X is not right (if you were to draw properly)

Holy crow. Now I don't need a ruler.

Great video. I just want to know what software or app do you use to make your video. Thanks

The drawing makes you believe that AB*-BB*=AB and AC*-CC*=AC. But in fact, if you draw with precision, you will notice that the correct thing is AB*-BB*=AB and AC-CC*=AC*. That changes all posterior conclusions. Nice trick!

AB = AB* - BB* but AC = AC* + CC* NOT (AC = AC* - CC*) The drawing is misleading......

i think the problem is that A X B* is not equal (Congruent) to A X C*

The error is in assuming that both B* and C* are exterior to the triangle, when in fact exactly one of them will be inside the triangle - the diagram is not drawn correctly. if we let B be the interior point, then the true relation of the side lengths is AB = AB* + BB* and AC = AC* - CC*.

1:16 is the mistake. M is the locus from B and C, but A's angular bisector would have to pass through M for the sides to be equal. Using this information we can deduce that AB does not equal AC

i tried it myself and C* came out to be nearer A than C so...

The fooling step is at 3'25". There is no reason to assume that the points C* and B* are at equal distance from A.

When I tried this in geogebra, XCC* was flipped so the right angle was nearer the the triangle. Angles were as follows: A = 78.08, B = 63.57, and C = 38.35 for anyone who wants to try.

Finally realized it and it's quite simple, the first thing to notice is that if this proof is wrong, then it must be hiding the error as some illegal construction as everything else after the construction is perfectly true axioms. Now the point is that you can never create the drawing shown in the video if you follow exact construction. C* will lie between A and C iff. AC>AB. Hence, even though CC* and BB* turn out to be equal, AC = AC* + C*C,while AB = AB* - B*B.

Upon first glance, this seems legit. I've rewatched this several times listening very closely to everything, and I realized something: If a triangle is equilateral, meaning it has all equal sides, that implies that same triangle also has equal ANGLES. This means that the angle bisector of angle A for the triangle you drew must also be the perpendicular bisector for line BC. That is clearly not the case in your example. But all other things check out, so I'm confused.

It's AXB* and AXC*, right? They are definitely not congruent.

If you did this on an equilateral triangle the diagram would hold true, but if either side is longer than the other the diagram will be different, like, if [side AC > side AB], CC* will partially be a part of the line AC. Hence, AC + CC* > AC* (as they have parts that will overlap).

*Triangles AB*X and AC*X aren't congruent (normally), because we **_only_** know that the hypotenuse (which is the side shared by the two triangles) and the shorter leg are the same, but only the angle opposite to the shorter and not the longer leg is equal. If the two triangles were congruent, they would form a rectangle which is clearly not happening normally.*

I love this video! Because although it's easy to see that the drawing is deliberately a little distorted, so that for example the angle at C* is obviously not 90°, it is very difficult to find any flaws in the logical arguments.

AXB* and AXC* are not congruent. The halfway point between B* and C* is on MX, not AX.

At 3:36 triangle AXB* is not congruent to AXC* because even if lengths B*X, XC* are equal, angles AB*X and AC*X are equal and angles XAB* and XAC* are equal, there is a difference between angles AXB* and AXC* which is not easy to see at first time because of the weak gap between the bisection and the bisector. That little difference implies that AB is different to AC owing to the non congruence of triangle AXB* and AXC*.

Point B* , point M and point C* are collinear, hence or B* is inside the triangle or C* is inside the triangle ( Simpson Line)

So... I guess all triangles are equilateral :D at least based on his diagram of course. I find no error in his proof (again based on the diagram). If you think SSA doesn't work, go look up HL theorem because we're dealing with right triangles. But the fault lies in the diagram itself. He drew the diagram not to scale on purpose. The angle bisector is really not an angle bisector. It's more evident in the red diagram towards the end of the video. In reality, the angle bisector and the perpendicular bisector would intersect much closer to the original triangle. Due to this, one of the * points will switch orders with the non-* point (either B and B* switch spots... or C and C* switch spots) Then the whole proof falls apart.

The error is the part from 4:31 through about 5:05, where he says you can do it again "on two other sides." But if you try that with, for example, angle C, A will lie on segment CA*, while B* will lie on segment CB, meaning that the final calculation beginning at 3:49 does not apply to the new construction. As others have pointed out, that calculation only works because the triangle is, in fact, isosceles as the proof shows, with AB = BC. That fact is deliberately obscured by a flawed drawing that deceitfully shows the angle bisector and the perpendicular bisector as separate when they should overlap. But the flaw in the drawing isn't the error. It just misdirects your attention away from the actual problem and gets you to spend hours trying to find a flaw where there is none. Without the flaw in the drawing, you would see that the proof, as far as it goes, looks correct and move on to the next logical step of rotating the triangle and seeing if it holds true for the other angles. You would quickly see that it does not, and the internet would miss out on quite a lot of silly people arguing for nonexistent flaws.

The two triangles XBB* and XCC* are NOT congruent Because we don't know if the two angles CXC* and BXB* are equal or not Note: we can say that two triangles are congruent if two sides of one triangle are equal to two sides of the other triangle, and the angle between them MUST be equal in both triangles. He didn't proof that the angles CXC* and BXB* are equal

If quadrilateral CABX is inscribed in a circle, which it can be, angles B and C must add up to 180 degrees, if line AX were a diameter, then the following configuration ensues where angle B = 90 degrees = angle C, and line AC is therefore equal to line AB. as triangle ABC is scalene, angle A =/= angle B, therefore one angle is obtuse and the other acute. Therefore, one, and only one point among either point B* or point C* which is among LINE AB or AC rather than a supposed ray AB or AC, hence disproving the method.

I think perpendicular bisector and angle bisector wouldn't intersect outside the triangle (I mean the right side of BC) (only their extensions would somewhere else)

There's nothing wrong with his reasoning, BUT the error comes from the fact that the point X is not drawn correctly. He drew the angle ever so slightly off so that the placement of X is incorrect. point X actually should be much closer to point M. The point C* should should be in between point A and point C. you can actually prove that its impossible to have both B* and C* outside of the triangle at the same time. one of them must be inside the triangle and one must be outside of the triangle. Angle Side Side IS valid if the angle is a right angle.

The two triangles XBB* and XCC* are NOT congruent as suggested, since same side-side-angle is NOT a property for congruency. Therefore, it's wrong to infer from this that BB* = CC*

This is a tricky problem indeed. One has to keep in mind that in order to solve such problems you must first draw a correct sketch. The error in this demonstration is in the assumption that point C* lies to the right of point C, when in fact it is situated LEFT of the point C.

4:06 is the bogus step.

The problem is simply that C* and B* cannot both lie outside of the triangle. Just use Simpson's Line.

If AB = AC, then a line that goes through the middle of BC and is also perpendicular to it, will also go through poing A. In that case there is no point X, but the whole line is X. Furthermore, point X is never outside the triangle, nor inside for that matter, but always on line BC. Then, if, say AB is longer than AC, point B* will be between points A and B, but point C will not be between points A and C. Hence, AB = AB*+BB*, while AC = AC*-CC* and if BB*=CC* and AB*=AC*, then it no longer follows that AB=AC.

It's because the BC Locus allows B and C to project to a point on the