Timeline for answer to Proving directly that, for scalene $\triangle ABC$, the bisector of $\angle A$ and perpendicular bisector of $BC$ meet at a point on the circumcircle by Nishant Kalonia
Current License: CC BY-SA 4.0
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12 events
| when toggle format | what | by | license | comment | |
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| May 25, 2025 at 11:15 | history | edited | Nishant Kalonia | CC BY-SA 4.0 |
Enhanced compactness
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| May 25, 2025 at 10:04 | vote | accept | Math12 | ||
| May 10, 2025 at 7:47 | history | edited | Nishant Kalonia | CC BY-SA 4.0 |
Specified Assumption
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| May 9, 2025 at 16:41 | history | edited | Nishant Kalonia | CC BY-SA 4.0 |
Added Diagram
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| May 8, 2025 at 16:51 | comment | added | Nishant Kalonia | @Math12 Since I knew that $A, B, C, K$ are going to be concyclic and $\triangle AOK$ isosceles, I realized that an all-angle approach would work, especially because of the elegant implication of the equality of two sides from the equality of two angles of an isosceles triangle. | |
| May 8, 2025 at 16:34 | comment | added | Math12 | Now, your proof seems to be correct. What was your intuition? How did you come up with the constructions and equations? | |
| May 8, 2025 at 14:54 | comment | added | Nishant Kalonia | I've changed the answer. | |
| May 8, 2025 at 14:54 | history | edited | Nishant Kalonia | CC BY-SA 4.0 |
Corrected the answer
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| May 8, 2025 at 14:27 | history | undeleted | Nishant Kalonia | ||
| May 8, 2025 at 13:56 | history | deleted | Nishant Kalonia | via Vote | |
| May 8, 2025 at 13:55 | comment | added | Math12 | You did the indirect approach. I explicitly said that I don't want to show $K=K'$. I want a direct alternate proof. | |
| May 8, 2025 at 13:51 | history | answered | Nishant Kalonia | CC BY-SA 4.0 |