msbiljanica Posted January 26, 2018 Report Share Posted January 26, 2018 load the attachment https://www.geogebra.org/m/KrYWuNEv see the description of the construction slider - $alpha$ -select the angle slider - point P - ruler with a socket, point Q must be line n Quote Link to comment Share on other sites More sharing options...
exchemist Posted January 26, 2018 Report Share Posted January 26, 2018 load the attachmenthttps://www.geogebra.org/m/KrYWuNEvsee the description of the constructionslider - $alpha$ -select the angleslider - point P - ruler with a socket, point Q must be line n Why? Quote Link to comment Share on other sites More sharing options...
Turtle Posted January 26, 2018 Report Share Posted January 26, 2018 The proportion of angels is always variable. :angel: Quote Link to comment Share on other sites More sharing options...
msbiljanica Posted January 28, 2018 Author Report Share Posted January 28, 2018 variable angle \alpha,constant ruler (PT\infty\infty)and divider (PUQ) , slider (P) is cross-sectional variable angle and constant ruler and divider ( point Q on the line n) Whether you are from the previous understand how it works proportions of angles, or that you explain step by step? Quote Link to comment Share on other sites More sharing options...
msbiljanica Posted February 7, 2018 Author Report Share Posted February 7, 2018 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 .... Quote Link to comment Share on other sites More sharing options...
Turtle Posted February 7, 2018 Report Share Posted February 7, 2018 Attachmentshttps://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 .... The classic trisection problem specifies tools as straightedge and compass only. Trisection using a ruler is not a problem. Quote Link to comment Share on other sites More sharing options...
msbiljanica Posted February 12, 2018 Author Report Share Posted February 12, 2018 Turtle - The classic trisection problem specifies tools as straightedge and compass only 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 Quote Link to comment Share on other sites More sharing options...
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