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 January 26th, 2018, 08:30 AM #1 Member   Joined: Jan 2015 From: leskovac Posts: 43 Thanks: 0 proportion of angles 2018 load the attachment https://www.geogebra.org/m/KrYWuNEv see the description of the construction slider - $\displaystyle \alpha$ -select the angle slider - point P - ruler with a socket, point Q must be line n
 January 27th, 2018, 11:41 PM #2 Member   Joined: Jan 2015 From: leskovac Posts: 43 Thanks: 0 Whether you are from the previous understand how it works proportions of angles, or that you explain step by step?
 January 28th, 2018, 09:50 AM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,091 Thanks: 795 What is your "point"?
 February 7th, 2018, 07:22 AM #4 Member   Joined: Jan 2015 From: leskovac Posts: 43 Thanks: 0 DENIS - the solution of the tricection of the angle Attachments https://www.geogebra.org/m/HQm7WwFk on the ruler $AB\infty{_1}\infty{_2}$ , raises divider ADC where AB + AB = AC, ruler sets the angle $\alpha$ semi-line ruler $B\infty{_1}$ sliding on point E , the point A of ruler slides semi-line l , when point C is on the line n , we get the radius of the circle , we get the angle $\beta$ we have solved the tricection of any angle Look at the construction protocol , or find the error if there is ....
 February 7th, 2018, 11:34 AM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,091 Thanks: 795 po_nt = point blank !
 February 12th, 2018, 07:37 AM #6 Member   Joined: Jan 2015 From: leskovac Posts: 43 Thanks: 0 https://www.geogebra.org/m/CukhmEVy straightedge slip on the point B - line i divider FIG , slides on straightedge , after slipping point F straighte line BC , point G describes lokus1 section lokus1 and line k point J , when changing the angle, the point J must be manually set to the intersection construction of regular polygon is possible by means of the proportion of the angle
April 13th, 2018, 03:21 AM   #7
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Quote:
 Originally Posted by point DENIS - the solution of the tricection of the angle Attachments https://www.geogebra.org/m/HQm7WwFk on the ruler $AB\infty{_1}\infty{_2}$ , raises divider ADC where AB + AB = AC, ruler sets the angle $\alpha$ semi-line ruler $B\infty{_1}$ sliding on point E , the point A of ruler slides semi-line l , when point C is on the line n , we get the radius of the circle , we get the angle $\beta$ we have solved the tricection of any angle Look at the construction protocol , or find the error if there is ....
Even if this has no error it is just another of the many ways of trisecting an angle using a sliding straight edge- and no one has ever said that was not possible.

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