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 Duong Ngo
Current License: CC BY-SA 4.0
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7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 10, 2025 at 6:09 | comment | added | Duong Ngo | Yes. Thanks for pointing that out. I fixed it. | |
| May 10, 2025 at 6:08 | history | edited | Duong Ngo | CC BY-SA 4.0 |
fixed typos
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| May 10, 2025 at 5:37 | comment | added | Math12 | In the solution, did you mean $\angle BEK=\angle BAK=\angle KAC$? | |
| May 9, 2025 at 13:56 | comment | added | Duong Ngo | I wanted to prove that $\angle KBA + \angle KCA = \pi$. So I thought about using congruent triangles. The law of sine was just a technical tool to work around because I couldn't show the triangles $KBE$ and $KCA$ are congruent. | |
| May 9, 2025 at 7:33 | comment | added | Math12 | How did come up with constructing E and using sine law? What was your intuition? | |
| May 8, 2025 at 15:07 | history | edited | Duong Ngo | CC BY-SA 4.0 |
a 2nd proof
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| May 8, 2025 at 14:36 | history | answered | Duong Ngo | CC BY-SA 4.0 |