Draw any two lines. Draw two lines connecting them. See figure at http://tinyurl.com/q6en8 From the line segments so formed, draw perpendicular bisectors. See figure at http://tinyurl.com/qu5yf From the intersection of the perpendicular bisectors, draw lines to the ends of the bisected line segments. See figure at http://tinyurl.com/rg9ur Six triangles are formed. (I labeled them with letters, but the letters are too small to see. They are A, B, C, D, E, and F, top to bottom and left to right. Triangles A and B are congruent by side, angle, side. Triangles E and F are congruent by side, angle, side. Triangles C and D are congruent by side, side, side. Label the angles at the common vertex with lower case corresponding to the capital letters. a = b; c = d; e = f, so a + c + e is equal to b + d + f a + b + c + d + e + f = = 2(a + b + c) = 360 degrees. a + b + c = 180 degrees. The two bisectors are evidently collinear and the two original lines, being perpendicular to it are parallel. Any other line may replace the one of the originals and the same proof applies. Therefore all lines are parallel. Q.E.D. Any comments? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
OT Paradox: all lines are parallel
Started by ●August 3, 2006
Reply by ●August 4, 20062006-08-04
> Triangles C and D are congruent by side, side, side.I don't think so. redraw your diagram with connecting lines very close to and very far away from the connection point of the original two lines. quo errat demonstrator :-)
Reply by ●August 4, 20062006-08-04
Jerry Avins skrev:> Draw any two lines. Draw two lines connecting them. > See figure at http://tinyurl.com/q6en8 > > From the line segments so formed, draw perpendicular bisectors. > See figure at http://tinyurl.com/qu5yf > > From the intersection of the perpendicular bisectors, draw lines to the > ends of the bisected line segments. > See figure at http://tinyurl.com/rg9ur > > Six triangles are formed. (I labeled them with letters, but the letters > are too small to see. They are A, B, C, D, E, and F, top to bottom and > left to right. > > Triangles A and B are congruent by side, angle, side. > Triangles E and F are congruent by side, angle, side. > Triangles C and D are congruent by side, side, side. > > Label the angles at the common vertex with lower case corresponding to > the capital letters. > > a = b; c = d; e = f, so a + c + e is equal to b + d + f > > a + b + c + d + e + f = = 2(a + b + c) = 360 degrees. > a + b + c = 180 degrees. > > The two bisectors are evidently collinear and the two original lines, > being perpendicular to it are parallel. > > Any other line may replace the one of the originals and the same proof > applies. > > Therefore all lines are parallel. Q.E.D. > > Any comments?The congruence arguments are only valid for right-angled triangles: Define a trianple PQR with R a right angle. From R, draw a line normal to PQ, denote the intersection with PQ as S. The triangles PQR and PRS are congruent. Your first line(!) asks for two *arbitrary* connecting lines to be drawn. Hence, the requirement for right-angled triangles is not met and the subsequent arguments based on congruence are invalid. Close? Rune
Reply by ●August 4, 20062006-08-04
stereo wrote:>> Triangles C and D are congruent by side, side, side. > > I don't think so. redraw your diagram with connecting lines very close > to and very far away from the connection point of the original two > lines. > > quo errat demonstrator :-)Right on! My geometry teacher (who also taught chemistry and was my faculty adviser) asked me to put a proof of something -- I forget what -- on the blackboard. I took what she thought was too long painstakingly drawing the figure. She said to get on with it, that it wasn't the figure that mattered, it was the proof. I was hurt, and I brooded about it all day. That night, I dreamed up the figure -- literally; I awoke at 3:AM and drew it on the pad I kept at my bedside for things I would forget by morning. In the morning I had to figure out what it meant. I had a 40 minute bus-and-subway ride to school, and by then I had the proof in my head (and the diagram to stare at). I drew the proof in a corner of the blackboard during the break before geometry. (We had 10 minutes between classes.) I told the teacher that there was evidently a flaw in the proof, but I couldn't find it. She looked at it quickly, found no obvious error. She had a class to teach, so she said she'd think about it at home. I gave her the paper with the diagram on it after I wrote the "proof" in the back. She began the next class and said that the "proof" depended on the perpendiculars intersecting between the lines. They don't. In fact, it was a good proof that they don't. She said that the flaw was in the diagram, not the logic. She told the class that my hoax showed that a careful diagram could be important. Then she said to me with a grin, "Good job". I was vindicated. That was one of life's little triumphs. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●August 4, 20062006-08-04
Rune Allnor wrote:> Jerry Avins skrev: >> Draw any two lines. Draw two lines connecting them. >> See figure at http://tinyurl.com/q6en8 >> >> From the line segments so formed, draw perpendicular bisectors. >> See figure at http://tinyurl.com/qu5yf >> >> From the intersection of the perpendicular bisectors, draw lines to the >> ends of the bisected line segments. >> See figure at http://tinyurl.com/rg9ur >> >> Six triangles are formed. (I labeled them with letters, but the letters >> are too small to see. They are A, B, C, D, E, and F, top to bottom and >> left to right. >> >> Triangles A and B are congruent by side, angle, side. >> Triangles E and F are congruent by side, angle, side. >> Triangles C and D are congruent by side, side, side. >> >> Label the angles at the common vertex with lower case corresponding to >> the capital letters. >> >> a = b; c = d; e = f, so a + c + e is equal to b + d + f >> >> a + b + c + d + e + f = = 2(a + b + c) = 360 degrees. >> a + b + c = 180 degrees. >> >> The two bisectors are evidently collinear and the two original lines, >> being perpendicular to it are parallel. >> >> Any other line may replace the one of the originals and the same proof >> applies. >> >> Therefore all lines are parallel. Q.E.D. >> >> Any comments? > > The congruence arguments are only valid for right-angled triangles: > Define a trianple PQR with R a right angle. From R, draw a line > normal to PQ, denote the intersection with PQ as S. The triangles > PQR and PRS are congruent. > > Your first line(!) asks for two *arbitrary* connecting lines to be > drawn. > Hence, the requirement for right-angled triangles is not met and the > subsequent arguments based on congruence are invalid. > > Close?No. Examine http://tinyurl.com/q6en8 . The two perpendicular bisectors are obviously at right angles to the segments. The two line . The other two line segments are equal by construction. There is no flaw in the proof. A triangles formed by a point on a perpendicular bisector to the ends of bisected line segments are congruent. Given the figure, the argument is valid, so the error must be in the figure. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●August 4, 20062006-08-04
Jerry Avins <jya@ieee.org> writes:> Then she said to me with a grin, "Good job". I was vindicated. That > was one of life's little triumphs.I think it also goes to show how much influence a teacher has on a young person. It's a real pity that our system (in the US) doesn't place the proper importance on educators. -- % Randy Yates % "The dreamer, the unwoken fool - %% Fuquay-Varina, NC % in dreams, no pain will kiss the brow..." %%% 919-577-9882 % %%%% <yates@ieee.org> % 'Eldorado Overture', *Eldorado*, ELO http://home.earthlink.net/~yatescr
Reply by ●August 4, 20062006-08-04
Randy Yates wrote:> Jerry Avins <jya@ieee.org> writes: > >> Then she said to me with a grin, "Good job". I was vindicated. That >> was one of life's little triumphs. > > I think it also goes to show how much influence a teacher has on a > young person. It's a real pity that our system (in the US) doesn't > place the proper importance on educators.My teacher was a refugee from Hitler's Germany. (She wasn't Jewish, and she got out early.) She had doctorates in math and chemistry (earned simultaneously). She knew Latin well and tutored Greek for those who wanted it. It was a small school, with classes of 25 or so. Education in Germany between the wars was evidently good too. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●August 4, 20062006-08-04
Jerry Avins <jya@ieee.org> writes:> Randy Yates wrote: >> Jerry Avins <jya@ieee.org> writes: >> >>> Then she said to me with a grin, "Good job". I was vindicated. That >>> was one of life's little triumphs. >> I think it also goes to show how much influence a teacher has on a >> young person. It's a real pity that our system (in the US) doesn't >> place the proper importance on educators. > > My teacher was a refugee from Hitler's Germany. (She wasn't Jewish, > and she got out early.) She had doctorates in math and chemistry > (earned simultaneously). She knew Latin well and tutored Greek for > those who wanted it. It was a small school, with classes of 25 or so. > > Education in Germany between the wars was evidently good too.Hi Jerry, Yours was a different era in the US. Things are (unfortunately) different now. Even back then, it sounds like you were very fortunate to have such a teacher. When I had geometry in 10th grade, we had an intern who taught part of the class. She took the time to plug through my proofs, which were somewhat unusual and long-winded, and give me feedback that "they worked." That was a great boost to my confidence. I had her again in 12th grade for "Honors Math Analysis" (essentially, first semester calculus). She was a great teacher - Mrs. Clare Merchant - and I owe her my gratitude. --Randy -- % Randy Yates % "Midnight, on the water... %% Fuquay-Varina, NC % I saw... the ocean's daughter." %%% 919-577-9882 % 'Can't Get It Out Of My Head' %%%% <yates@ieee.org> % *El Dorado*, Electric Light Orchestra http://home.earthlink.net/~yatescr
Reply by ●August 7, 20062006-08-07
Randy Yates wrote: (snip)> When I had geometry in 10th grade, we had an intern who taught > part of the class. She took the time to plug through my proofs, > which were somewhat unusual and long-winded, and give me feedback > that "they worked." That was a great boost to my confidence.My 10th grade geometry teacher's description of our proofs, "Why do it the easy way when there's a hard way." Why is it that we remember our geometry class so well? -- glen
Reply by ●August 8, 20062006-08-08
glen herrmannsfeldt wrote: ...> Why is it that we remember our geometry class so well?Elementary arithmetic is trivial in that is just something that one does. Geometry requires invention -- every proof is either an invention or a copy of one -- and exhibits and exploits the relations between things. Other subjects exhibit and explore relations, but geometry is unique among them in every conclusion's being either correct or not. It requires thought, analysis, and planning a sequence of steps, but unlike other subjects with those requirements, it isn't fuzzy. No points of view are involved, no matters of opinion. For some students, that's a drag. For others -- most of us here, I'll wager -- it is an epiphany, a realization that we can think critically, and that thinking matters. Of course we remember it. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
