Timeline for ten concurrent lines
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 2, 2014 at 23:39 | vote | accept | abel | ||
| Dec 1, 2014 at 21:19 | comment | added | Aaron Meyerowitz | @abel One way to spin it is that the result follows from "in the same direction as the line from the origin to $d+e$" So you can say it just like that and not require equal lengths -OR- do require that and say "in the direction perpendicular to the cord $de$ which, because we required that all lengths are equal, is the same direction as the line from the origin to $d+e$." | |
| Dec 1, 2014 at 21:02 | comment | added | abel | don't you still need the lengths of $d$ and $e$ to equal so that $d+e$ and $d-e$ are orthogonal. that all five of points are on the circle is still needed. | |
| Dec 1, 2014 at 20:55 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
added 2446 characters in body
|
| Dec 1, 2014 at 19:38 | comment | added | abel | i like the vector proof. but does this not assume the form for the common point of intersection. even in my earlier proof, i could have argued that by symmetry the form must be as claimed. | |
| Dec 1, 2014 at 19:28 | history | answered | Aaron Meyerowitz | CC BY-SA 3.0 |