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January 26th, 2018, 09:30 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 28th, 2018, 12:41 AM  #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, 10:50 AM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,799 Thanks: 970 
What is your "point"?

February 7th, 2018, 08:22 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 $\alpha$ semiline ruler $B\infty{_1}$ sliding on point E , the point A of ruler slides semiline 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, 12:34 PM  #5 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,799 Thanks: 970 
po_nt = point blank !

February 12th, 2018, 08:37 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, 04:21 AM  #7  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895  Quote:
 

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