Pons asinorum: what's wrong with this proof?
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From: kj on 6 Aug 2010 16:31 According to Wikipedia, the popular explanation for the nickname "Pons asinorum" given to one of Euclid's theorem is that it was considered "the first real test ... of the intelligence of the reader. The Wikipedia also shows a graphic version of Euclid's proof, by Byrne. This proof seems to me insanely complex when compared to the following proof. Let ABC be an isosceles triangle, with BA = BC. We want to show that angles BAC and BCA are equal. Proof: Produce a second triangle C'B'A' by reflecting ABC relative to the line passing through B and perpendicular to AC (actually, any line will do). Since BA = BC = B'C' = B'A', and AC = C'A', it follows that ABC and C'B'A' are congruent. Therefore angles BAC and B'C'A' are equal. But B'C'A' is just the reflection of BCA. Therefore angle BAC = angle BCA. QED. I suppose that the flaw in this proof is that it relies on the fact that angles and lengths are invariant with respect to reflection, which may be obvious but had not been proven yet by that point in Euclid's book. (???) ~K
From: Chip Eastham on 6 Aug 2010 19:22 On Aug 6, 4:31 pm, kj <no.em...(a)please.post> wrote: > According to Wikipedia, the popular explanation for the nickname > "Pons asinorum" given to one of Euclid's theorem is that it was > considered "the first real test ... of the intelligence of the > reader. The Wikipedia also shows a graphic version of Euclid's > proof, by Byrne. > > This proof seems to me insanely complex when compared to the > following proof. > > Let ABC be an isosceles triangle, with BA = BC. We want to show > that angles BAC and BCA are equal. > > Proof: Produce a second triangle C'B'A' by reflecting ABC relative > to the line passing through B and perpendicular to AC (actually, > any line will do). Since BA = BC = B'C' = B'A', and AC = C'A', it > follows that ABC and C'B'A' are congruent. Therefore angles BAC > and B'C'A' are equal. But B'C'A' is just the reflection of BCA. > Therefore angle BAC = angle BCA. QED. > > I suppose that the flaw in this proof is that it relies on the fact > that angles and lengths are invariant with respect to reflection, > which may be obvious but had not been proven yet by that point in > Euclid's book. (???) > > ~K It's hard to tell whether you are asking about the meaning of the phrase "pons asinorum" or for comments on the validity of your(?) proposed simpler proof. "Pons" here is Latin for bridge, and Euclid's diagram for the isoceles triangle, with its equal sides extended beyond the base, looks something like a bridge. Certainly the cited proposition (fifth in Book 1 of Euclid) appears with little "machinery" yet built in the way of prior proofs. Euclidean axioms provide for the ability of line segments to be extended indefinitely, but there's no axiomatic notion of reflecting a triangle in a line. Your suggestion that the proposed proof is out of place with respect to the incremental proof of results in Euclid seems most apt: Prop. 1: An equilateral triangle can be constructed with a given side. Prop. 2: A line segment ending at point A can be constructed equal to a given BC. Prop. 3: Given two unequal line segments, the greater of these can have a part cut off that is equal to the lesser of these. Prop. 4: If two triangles have two pairs of respective sides equal, and if the respective angles between those sides are also equal, the two triangles have all respective sides and angles equal (congruence). regards, chip
From: kj on 6 Aug 2010 21:15 In <9b114bbe-0cfb-40a8-95a7-68a11a0df650(a)v41g2000yqv.googlegroups.com> Chip Eastham <hardmath(a)gmail.com> writes: >On Aug 6, 4:31 pm, kj <no.em...(a)please.post> wrote: >> According to Wikipedia, the popular explanation for the nickname >> "Pons asinorum" given to one of Euclid's theorem is that it was >> considered "the first real test ... of the intelligence of the >> reader. The Wikipedia also shows a graphic version of Euclid's >> proof, by Byrne. >> >> This proof seems to me insanely complex when compared to the >> following proof. >> >> Let ABC be an isosceles triangle, with BA = BC. We want to show >> that angles BAC and BCA are equal. >> >> Proof: Produce a second triangle C'B'A' by reflecting ABC relative >> to the line passing through B and perpendicular to AC (actually, >> any line will do). Since BA = BC = B'C' = B'A', and AC = C'A', it >> follows that ABC and C'B'A' are congruent. Therefore angles BAC >> and B'C'A' are equal. But B'C'A' is just the reflection of BCA. >> Therefore angle BAC = angle BCA. QED. >> >> I suppose that the flaw in this proof is that it relies on the fact >> that angles and lengths are invariant with respect to reflection, >> which may be obvious but had not been proven yet by that point in >> Euclid's book. (???) >> >> ~K >It's hard to tell whether you are asking about >the meaning of the phrase "pons asinorum" or >for comments on the validity of your(?) proposed >simpler proof. Agreed, my question was not very clear. What I was trying to get at was that if Euclid's proof of the proposition was the "first real test ... of the intelligence of the reader" it was so only due to its unnecessary complexity (as it seemed to me), since a *much* simpler and obvious proof seemed to me possible. Unless, of course, this simpler proof was wrong somehow, which would in itself make the complexity of Euclid's proof seem less out-of-place. But your clarifications explained the situation perfectly. Thanks! ~K
From: bert on 7 Aug 2010 07:14 On 6 Aug, 21:31, kj <no.em...(a)please.post> wrote: > According to Wikipedia, the popular explanation for the nickname > "Pons asinorum" given to one of Euclid's theorem is that it was > considered "the first real test ... of the intelligence of the > reader. The Wikipedia also shows a graphic version of Euclid's > proof, by Byrne. > > This proof seems to me insanely complex when compared to the > following proof. > > Let ABC be an isosceles triangle, with BA = BC. We want to show > that angles BAC and BCA are equal. > > Proof: Produce a second triangle C'B'A' by reflecting ABC relative > to the line passing through B and perpendicular to AC (actually, > any line will do). Since BA = BC = B'C' = B'A', and AC = C'A', it > follows that ABC and C'B'A' are congruent. Therefore angles BAC > and B'C'A' are equal. But B'C'A' is just the reflection of BCA. > Therefore angle BAC = angle BCA. QED. There is an anecdote about a very early "artificial intelligence" program, set up to construct geometrical proofs starting from Euclid's axioms. It crossed this bridge by proving that the triangle ABC is congruent to the triangle CBA (by two sides and their included angle), hence the angles at A and C are equal. Euclid seems to have been happy with invariance under reflection, at least to the extent that his rules allow an arbitrary triangle to be proved congruent to its own mirror image. --
From: Chip Eastham on 7 Aug 2010 10:19
On Aug 7, 7:14 am, bert <bert.hutchi...(a)btinternet.com> wrote: > On 6 Aug, 21:31, kj <no.em...(a)please.post> wrote: > > > > > According to Wikipedia, the popular explanation for the nickname > > "Pons asinorum" given to one of Euclid's theorem is that it was > > considered "the first real test ... of the intelligence of the > > reader. The Wikipedia also shows a graphic version of Euclid's > > proof, by Byrne. > > > This proof seems to me insanely complex when compared to the > > following proof. > > > Let ABC be an isosceles triangle, with BA = BC. We want to show > > that angles BAC and BCA are equal. > > > Proof: Produce a second triangle C'B'A' by reflecting ABC relative > > to the line passing through B and perpendicular to AC (actually, > > any line will do). Since BA = BC = B'C' = B'A', and AC = C'A', it > > follows that ABC and C'B'A' are congruent. Therefore angles BAC > > and B'C'A' are equal. But B'C'A' is just the reflection of BCA. > > Therefore angle BAC = angle BCA. QED. > > There is an anecdote about a very early > "artificial intelligence" program, set up > to construct geometrical proofs starting > from Euclid's axioms. It crossed this > bridge by proving that the triangle ABC is > congruent to the triangle CBA (by two sides > and their included angle), hence the angles > at A and C are equal. > > Euclid seems to have been happy with invariance > under reflection, at least to the extent that > his rules allow an arbitrary triangle to be > proved congruent to its own mirror image. > -- A good account of Euclid's proof and analysis of why there are no "shortcuts" to the result is here: http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI5.html In proving the "interior" base angles of the isoceles triangle are equal, Euclid can and does make use of SAS congruence (of the triangle to itself, in essence) just proven in Prop. 4 of Book I. However Prop. 5 has a second conclusion, that "exterior" base angles, the result of extending the equal sides beyond the base, are also equal. It "proves" to be a good bit more subtle to establish. regards, chip |