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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 rig
by 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 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.