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Proving that a parallelogram ABCD with equal diagonals is a rectangle is an exercise, but not too hard. You can make two different triangles ABC and ABD with th

Proving that a parallelogram ABCD with equal diagonals is a rectangle is an exercise, but not too hard. You can make two different triangles ABC and ABD with the same parallelogram base, where one has second side BC and third side (a diagonal) AC, and the other has second side AD and third side (the other diagonal) BD. These triangles are SSS congruent, therefore must have congruent angles (Euclid I.8) ABC and BAD. This gives you a parallelogram with 4 congruent angles.

muro 2y ago
Nice. You can alternatively prove the negative: suppose it's a parallelogram with non-right angles, assume angle at A is "sharper". Thus, you can create two right angle triangles: AB'C with the base extended beyond B (= B') where AC is one diagonal and A'BD where A' is "below" D and BD is the other diagonal. As AB'C is larger than A'BD (because A' is within AB and B' is outside while the height is the same) AC can't be the same length as BD. Thus, no parallelogram with non-right angles exists that has equal length diagonals.
- 3abiton 2y ago
  I miss this kind of maths. Any book recommendations for such maths puzzles/questions?
  - muro 2y ago
    +1, would also enjoy a book recommendation. I can recommend a fun app - Euclidea :) My favorite book on the topic, unfortunately only in Slovak, is: "Matematici, ja a ty" (mathematicians, you and me). I don't think there is a translation. I also had a university professor who had a similar style - not dry proofs, but "Here is how Archimedes thought about proofs" - which is related to the comment above: if A is not smaller than B and B is not smaller than A, they need to be equal.
akho 2y ago
Sure. The overall proof of Thales is then about as complicated as the “standard” one with two isosceles triangles.
- jacobolus 2y ago
  Breaking proofs of trickier theorems up into a few parts and hiding the details of each part behind another theorem whose proof is self-contained and easy to follow (even better if it's "obviously true") is the basic idea of mathematics. Trying to make the top-level proof very clearly demonstrate, without a lot of bookkeeping or head scratching, that the theorem must be true is a wonderful goal. In my opinion it's a lot more obvious that a parallelogram with 2 equal diagonals must be a rectangle than that an inscribed angle intersecting a circle at ends of a diameter must be a right angle.
  - akho 2y ago
    Whether this proof is complete depends on what was known before. Was the lemma introduced by a teacher? was it not actually proven, and the student missed the hole in their proof? these are different teachable moments. I think the two facts, by themselves, are roughly equivalent in difficulty. Moving from Thales to equal diagonals parallelogram requires drawing an extra circle; moving from parallelogram facts to Thales requires either the construction in the article, or continuing the missing radius to get the same rectangle.
    - jacobolus 2y ago
      I don't understand why you are invested in nitpicking the work of an anonymous 7th grader from decades ago. Clearly, filling in all of the details of any formal proof in Greek style is going to require a careful knowledge of the axioms in use and some list of previously proven theorems which are allowed to be used without re-proof. Depending on which theorems are at hand already, one or another proof might be shorter or longer or more or less obvious. When trying to construct a whole mathematical treatise, the best order for the theorems so that each proof depends only on previously proven theorems is a tricky choice involving some trade-offs between pedagogical goals and concision of individual proofs. But all of that is missing the point.
akho 2y ago
In the spirit of the posted proof, noting axial symmetry would be neater. (But then you need to know that parallelogram diagonals split each other in half, …)
- deleted 2y ago
  [deleted]

muro 2y ago

Nice. You can alternatively prove the negative: suppose it's a parallelogram with non-right angles, assume angle at A is "sharper". Thus, you can create two right angle triangles: AB'C with the base extended beyond B (= B') where AC is one diagonal and A'BD where A' is "below" D and BD is the other diagonal. As AB'C is larger than A'BD (because A' is within AB and B' is outside while the height is the same) AC can't be the same length as BD. Thus, no parallelogram with non-right angles exists that has equal length diagonals.

3abiton 2y ago
I miss this kind of maths. Any book recommendations for such maths puzzles/questions?
- muro 2y ago
  +1, would also enjoy a book recommendation. I can recommend a fun app - Euclidea :) My favorite book on the topic, unfortunately only in Slovak, is: "Matematici, ja a ty" (mathematicians, you and me). I don't think there is a translation. I also had a university professor who had a similar style - not dry proofs, but "Here is how Archimedes thought about proofs" - which is related to the comment above: if A is not smaller than B and B is not smaller than A, they need to be equal.

akho 2y ago

Sure. The overall proof of Thales is then about as complicated as the “standard” one with two isosceles triangles.

jacobolus 2y ago
Breaking proofs of trickier theorems up into a few parts and hiding the details of each part behind another theorem whose proof is self-contained and easy to follow (even better if it's "obviously true") is the basic idea of mathematics. Trying to make the top-level proof very clearly demonstrate, without a lot of bookkeeping or head scratching, that the theorem must be true is a wonderful goal. In my opinion it's a lot more obvious that a parallelogram with 2 equal diagonals must be a rectangle than that an inscribed angle intersecting a circle at ends of a diameter must be a right angle.
- akho 2y ago
  Whether this proof is complete depends on what was known before. Was the lemma introduced by a teacher? was it not actually proven, and the student missed the hole in their proof? these are different teachable moments. I think the two facts, by themselves, are roughly equivalent in difficulty. Moving from Thales to equal diagonals parallelogram requires drawing an extra circle; moving from parallelogram facts to Thales requires either the construction in the article, or continuing the missing radius to get the same rectangle.
  - jacobolus 2y ago
    I don't understand why you are invested in nitpicking the work of an anonymous 7th grader from decades ago. Clearly, filling in all of the details of any formal proof in Greek style is going to require a careful knowledge of the axioms in use and some list of previously proven theorems which are allowed to be used without re-proof. Depending on which theorems are at hand already, one or another proof might be shorter or longer or more or less obvious. When trying to construct a whole mathematical treatise, the best order for the theorems so that each proof depends only on previously proven theorems is a tricky choice involving some trade-offs between pedagogical goals and concision of individual proofs. But all of that is missing the point.

akho 2y ago

In the spirit of the posted proof, noting axial symmetry would be neater. (But then you need to know that parallelogram diagonals split each other in half, …)

deleted 2y ago
[deleted]