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Proving that a parallelogram with equal diagonals is a rectangle is an exercise in itself; I’d prove it through Thales’ theorem, myself... Also, Lockhart's Lam
by akho 2y ago
Proving that a parallelogram with equal diagonals is a rectangle is an exercise in itself; I’d prove it through Thales’ theorem, myself...
Also, Lockhart's Lament is from 2002, so this post probably needs a (2002). It is very unlikely that the proof was new. It was certainly new to the seventh-grader, and a great result at that.
- lloeki 2y ago> Proving that a parallelogram with equal diagonals is a rectangle is an exercise in itself; I’d prove it through Thales’ theorem, myself.. IIRC you prove it via angles. a) The sum of angles of any triangle is pi. b) The sum of angles of any quadrilateral is 2*pi. c) Since you rotate the triangle, opposite angles end up adding together. Because of the above, a bit of reasoning around symmetry shows that when diagonals are equal they can only add to pi/2 and that the other ones can only be pi/2. Any other angle leads to a contradiction. I mentioned angle values but (again IIRC) this can all be proven with compass and ruler.
- vidarh 2y agoAs the article mentions, in this particular case it's simpler: You have a point on the triangle that is also on the circle. Now you rotate the triangle 180 degrees, with the point following the circle. The point inherently must end up exactly opposite the original point, and so the diagonal formed between the old and new point must be the diameter of the circle.
- lloeki 2y agoThis proves that the parallelogram diagonals, both being diameters of the same circle, are equal, akho was pointing out that there's a leap from that to "the parallelogram is a rectangle", which can be proven with Thales. But then, the article is about laying out a proof of Thales so if it's using a lemma that leveraged Thales itself that's circular logic right there and so we're in trouble. One other way to prove Thales is by computing angles, but then you've got yourself a proven Thales so the point of the article is moot.
- vidarh 2y agoI didn't read that very carefully, but proving that a parallelogram with equal diagonals is a rectangle is obvious enough via triangle congruency that I suspect they were just taught it without the proof. It's not a particularly interesting detail.
- vlovich123 2y agoUm… small correction. Sum of angles in triangle is 180 not pi (right-angle triangle is 90 + 45 + 45). Similarly for quadrilateral it’s 360 not 180 (a canonical being rectangles which is 4 90 degree angles).
- akho 2y agoWhat do you think pi is.
- vlovich123 2y agoApparently HN doesn't support the facepalm emoji.
- deleted 2y ago[deleted]
- akho 2y ago“Standard” Thales proof (via two isosceles triangles) is simpler than this.
- lloeki 2y agoIt is, but Thales is what this article attempts to prove (a). By the point you've proven Thales with isosceles (b), proving that a parallelogram with equal diagonals is a rectangle thanks to Thales(b) in order to go forward along the Thales(a) proof is circular logic.
- jasomill 2y agoIn case (a) and (b) aren't immediately obvious: The net change in direction around a triangle, quadrilateral, or any simple closed curve is a full rotation, i.e., 2π. The sum of the interior angle and the signed change in direction around a vertex of a polygon is π (consider the case of a "vertex" with interior angle π). So for a polygon with interior angles θ₁, θ₂, …, θₙ, 2π = (π - θ₁) + (π - θ₂) + ⋯ + (π - θₙ), therefore θ₁ + θ₂ + ⋯ + θₙ = nπ - 2π = (n-2)π.
- jacobolus 2y agoProving 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 agoNice. 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 agoI 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 agoSure. The overall proof of Thales is then about as complicated as the “standard” one with two isosceles triangles.
- deleted 2y ago[deleted]
- tmaly 2y agoI noticed the link on the link to Paul's post was not working.
- dang 2y agoChanged from 2008 to 2002 above. Thanks!