Timeline for Tetrahedron angles sum to $\pi$: Bisector plane
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jul 30, 2017 at 10:54 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
deleted 21 characters in body; edited tags
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| Apr 29, 2012 at 13:46 | vote | accept | Joseph O'Rourke | ||
| Apr 20, 2012 at 14:41 | answer | added | Lee Mosher | timeline score: 3 | |
| Apr 20, 2012 at 11:58 | comment | added | Misha | @Joseph: Take $a'$ obtained by reflecting $a$ in the point $b$ (central symmetry) and dilating with center $b$ so that $|ab'|=|cb|$. Then you are asking for $d$ so that $\angle dba'=\angle dbc$ (since $a, b, a', d$ are coplanar. Since the triangles $\Delta dba'$ and $\Delta dbc$ are now congruent (two sides and the angle between them), you are asking for $|da'|=|dc|$, i.e. $d$ is on the bisector plane of $a', c$, which is exactly your claim. Of course, this is very similar to the arguments of both fedja's and Gjergji's. | |
| Apr 20, 2012 at 1:55 | answer | added | Gjergji Zaimi | timeline score: 5 | |
| Apr 20, 2012 at 1:03 | answer | added | fedja | timeline score: 4 | |
| Apr 20, 2012 at 0:49 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |