ODT2: All triangles are isosceles
Voting Style:  Open  Point System:  7 Point  
Started:  11/10/2012  Category:  Science  
Updated:  5 years ago  Status:  Post Voting Period  
Viewed:  5,361 times  Debate No:  27092 
This debate is part of Official Debate Tournament 2 (ODT2). I will present a proof that triangles are isosceles. Clearly all triangles are not isosceles. There must be a flaw in my proof. The task of my opponent is to spot the flaw. == Structure of Debate == R1 is for acceptance. In R2 I will present my proof. My esteemed opponent can explain the flaw in R2, R3 or R4. If she manages to spot the flaw, she wins the debate. In case she fails in her task, I will explain the flaw in R5 and claim victory. Pro need not do anything in R2, R3 and R4. Con does not have to do anything in R5. Con has promised to work on the problem alone. More specifically, she has agreed to not to search internet or to ask friends for help. Voters will not vote based on truth of resolution (it is false). Voters will vote Pro if my opponent fails to spot the mistake in my proof. Otherwise they will vote Con.
I accept the terms of the debate. 

I would like to thank Royalpaladin for prompt acceptance of the challenge (within few minutes). == To Prove == All Triangles are isosceles Consider an arbitrary triangle ΔABC, shown in the figure below. Our goal is to show that AB = AC. This should be sufficient to establish that all triangles are isosceles. == Construction ==
The resulting diagram should look as follows. == Claim 1 == ΔAOE ≜ ΔAOF Proof: ∠OEA = ∠OFA ( = 90^{0} ) ∠EAO = ∠FAO (AO is angle bisector of ∠A) AO = AO (Common) By AAS congruency, ΔAOE ≜ ΔAOF == Claim 2 == ΔBOD ≜ ΔCOD Proof: ∠BDO = ∠CDO ( =90^{0} ) DO = DO (Common) By SAS congruency, ΔBOD ≜ ΔCOD == Claim 3 == ΔEOB ≜ ΔFOC Proof: ∠OEB = ∠OFC( =90^{0} ) BO = CO (ΔBOD ≜ ΔCOD as per claim 2) EO = FO (ΔAOE ≜ ΔAOF as per claim 1) By RHS congruency, ΔEOB ≜ ΔFOC == Claim 4 == AB = AC Proof: AE = AF(ΔAOE ≜ ΔAOF as per claim 1) EB = FC (ΔEOB ≜ ΔFOC as per claim 3) AE + EB = AF + FC(Adding last two lines) AB = AC Claim 4 establishes most of the proof. There are few more scenarios we need to consider before we can call the proof to be complete. <<<< Alternative Scenario 1 >>>> In construction step 1, we assumed that angle bisector of ∠A and perpendicular bisector of side BC intersect each other. Consider the situation where angle bisector of side A is perpendicular to side BC and intersects BC and D’. On other hand, perpendicular bisector of side BC is DA’.
Consider ΔAD’B and ΔAD’C AD’ = AD’ (Common) ∠AD’B = ∠AD’C ( =90^{0} ) From ASA congruency, ΔAD’B ≜ ΔAD’C This means, AD’ = D’C. This implies D’ is midpoint of BC. However midpoint of BC was assumed to be D (as A’D was perpendicular bisector). This means D and D’ are identical. This also means AD’ and AD are identical lines. Most important of all, AB = AC. This triangle is isosceles. <<<< Alternative Scenario 2 >>>> What if the angle bisector and perpendicular bisector are almost parallel? In this case point O will lie below the line BC. I could have asked reader to complete this part of the proof, however I will do it myself for sake of completeness. == Construction ==
In this case the perpendicular dropped from O on line AB and BC will lie on extended part of the respective line. Apart from that, all the constructions are identical to previous cases. == Claim 1 == ΔAOE ≜ ΔAOF Proof: ∠OEA = ∠OFA ( = 90^{0} ) ∠EAO = ∠FAO (AO is angle bisector of ∠A) AO = AO (Common) By AAS congruency, ΔAOE ≜ ΔAOF == Claim 2 == ΔBOD ≜ ΔCOD Proof: BD = CD(OD is perpendicular bisector of AD) ∠BDO = ∠CDO ( =90^{0} ) DO = DO (Common) By SAS congruency, ΔBOD ≜ ΔCOD == Claim 3 == ΔEOB ≜ ΔFOC Proof: ∠OEB = ∠OFC ( =90^{0} ) BO = CO(ΔBOD ≜ ΔCOD as per claim 2) EO = FO (ΔAOE ≜ ΔAOF as per claim 1) By RHS congruency, ΔEOB ≜ ΔFOC == Claim 4 == AB = AC Proof: AE = AF(ΔAOE ≜ ΔAOF as per claim 1) EB = FC(ΔEOB ≜ ΔFOC as per claim 3) AE  EB = AF  FC (From last two lines) AB = AC This completes the proof. == Corollary == All triangles are equilateral. We have proved AB = AC. Using similar procedures we can show that AB = BC. This would mean AB = BC = CA. Thus all triangles are equilateral. Over to my esteemed opponent.
I have not had much time to work on this. I will wait to post the flaw until next round. 

Quick Note My opponent must prove that *all* triangles are isosceles. If I can show that even one triangle is isosceles, I win the round because then not *all* triangles are isosceles. Definition of Isosceles Triangle and Impact An isosceles triangle is a triangle with at least two sides that are equal. The 345 right triangle is not an isosceles triangle, so we know that my opponent’s case is flawed from the outset. Opponent’s Case My opponent attempts to prove that all triangles are isosceles through an examination of three different cases that center around the intersection of an angle bisector and the perpendicular bisector of the side across from it. It is a given that if the perpendicular bisector and the angle bisector are the same, the triangle will be isosceles, so he needs to prove the three other cases. Scenario 1: The perpendicular bisector and the angle bisector intersect in the middle of the triangle. This one gave me a great deal of trouble, but after staring at the proof for two hours, I noticed that my opponent’s diagram is actually misleading and then did some investigation of my own through drawings on paper. I found that Scenario 1 does not even exist since the perpendicular bisector and the angle bisector will always intersect outside the triangle. This proof hinges on an internal intersection, but that does not happen. Scenario 1 thus turns into Scenario 3. Scenario 2: No intersection This scenario doesn’t actually make sense; if you look at my opponent’s drawings, the triangles cannot possibly be congruent because one has larger sides than the other. The flaw in this scenario is that he assumes that the angle bisector will intersect the side at a 90 angle, which is not possible if it is not equal to the perpendicular bisector. In fact, the two subtriangles could not ever be congruent (which requires all sides and angles to be the same) if the angle bisector and perpendicular bisector attached to the side at two different points since the angle bisector would not divide the triangle into equal halves. Scenario 3: External Intersection This is where my opponent does not actually prove his case because he does not consider a scenario that actually is part of this scenario. My opponent is correct that lines can be drawn at OB and OC, but he then tells you that he wants to connect O and AC and AB. He assumes that a ray from O will intersect these lines outside of the triangle, but it is possible to have a situation in which E, for example, is between A and B. In this case, OE and OF would not be congruent, and neither would AOE and AOF. This was claim one in his proof, so in this scenario, the triangle is not isosceles and the resolution is negated. 

== Quick notes on quick notes == This one gave me a great deal of trouble, but after staring at the proof for two hours… Glad to know that my opponent enjoyed the problem. I am glad to note that she has made some progress. In fact she has spotted where the flaw lies. However her analysis of the problem is still incomplete. An isosceles triangle is a triangle with at least two sides that are equal. The 345 right triangle is not an isosceles triangle, so we know that my opponent’s case is flawed from the outset. No doubt the proof is flawed. However the task of Con is to find the flaw. Inside Intersection: My opponent thinks this scenario can never happen because… …did some investigation of my own through drawings on paper. I found that Scenario 1 does not even exist since the perpendicular bisector and the angle bisector will always intersect outside the triangle. There are two obvious problems with the approach.
No intersection: My opponent has argued that this case can never occur. I agree. She should read my proof again. I have also proved the same case by showing D and D’ must coincide. If they coincide this is the case for isosceles triangle. External Intersection: She has correctly spotted that I have skipped few cases. Her analysis of the missing case is wrong. I guess she will do a more thorough (and formal) job in next round. To win this debate, the least we can expect from her is to post the correct analysis of the missing scenario and explain why the proof breaks down in that case. It shall be unfortunate if she restricts herself to doing the minimum for winning the debate. It will be intellectually satisfying if she can also explain the flaw in scenario where both E and F lie outside the triangle. (Apart from the internal intersection scenario). Wishing my opponent a few more hours of staring at the proof… Notes I am going to go ahead and protest my opponent’s claim that I need to provide a formal proof for my claims. In Round 1, he explicitly said that I just needed to find the flaw in his proof, and in the previous round, he admits that I found the flaw. This means that I have already won the debate. However, in case some of you disagree, I will continue. Note, however, that by the rules he established in the first round, you should vote Con without reading the rest of this debate. Scenario 1 He asks me to prove that this scenario cannot occur and notes that my test cases don’t prove anything. Fine. I will ask the readers to construct the diagrams at home since I cannot do so for them. A theorem known as the Internal Angle Bisector Theorem notes that if there is a triangle ABC with an internal angle bisector AD of the angle BAC, AB/AC=BD/DC (the ratios will be the same). Now, suppose we have a triangle ABC with an internal angle bisector D and a perpendicular bisector of BC at point M. In order to prove my claim true, I need to demonstrate that BD<BM (since if BD was greater, they would intersect inside the triangle). BC=BM+MC (since M is the midpoint) BC=BD+DC BD<DC Proof that BD<BC: By the Internal Angle Bisector Theorem AB/AC=BD/DC However, we know that AB is less than AC Thus, AB/AC <1 If AB/AC<1 and BD/DC=AB/BC, BD/DC<1 and BD<DC 2BM=BM+MC=BD+DC>BD+BD BD<BM Thus, Scenario 1 never occurs. Scenario 2 I already conceded that Scenario 2 only occurs if D and D’ are the same, in which case the triangle is isosceles. Otherwise, the scenario does not occur. Scenario 3 Note that I only need to prove that in the case in which E is not outside the triangle, the triangle is not isosceles in order to win. So, if E is between A and B and F is outside the triangle and on line AC, the proof breaks down. He was right; my previous argument was incorrect. His proof actually breaks down at Claim 4 in this scenario because he says that AE  EB = AF  FC. This is false in the case I presented; AEEB not equal AFFC when E is between A and B. This means that AB will not equal AC and the triangle is not isosceles. I don’t need to prove the other scenario false since I have shown that not all triangles are isosceles. 

My opponent has correctly spotted the flaw. On other hand her insistence that the readers draw there own diagram at home appears to be weird. VOTE CON This problem appears in 'Fallacies in Mathematics' by EA Maxwel [1]. The proof is often attributed to Charles Dodgeson who is better known for his novel Alice Adventures in Wonderland by Lewis Carrol). [1] http://www.scribd.com...
I would like to thank my opponent for a good debate. As he said, vote Con. 
baggins  royalpaladin  Tied  

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baggins  royalpaladin  Tied  

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baggins  royalpaladin  Tied  

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baggins  royalpaladin  Tied  

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baggins  royalpaladin  Tied  

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baggins  royalpaladin  Tied  

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baggins  royalpaladin  Tied  

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baggins  royalpaladin  Tied  

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You guys probably kept the confusing to the voters, because after, they were still undecided, since again they all voted tie.
if yes then you are really a genius
Redefining the metric would have been a nice trick. However this proof is constrained to high school geometry (i.e R2 with the usual metric).
By the way, since there is no reference to the structure of the space where the triangles lie, one could assume they are on R^2 with the trivial metric d((x,x'),(y,y'))=1, for all real x,x',y,y'. Hence all triangles are isosceles and furthermore they are equilateral.
I guess Pro assumes they are in R^2 with the usual metric d((x,x'),(y,y'))=sqrt((xy)^2+(x'y')^2).
*looks at comments*
*repeats Maikuru's comment*
+ 1000
done.