1. ## proportioned angles

Required accessories - pencil, compass, unmarked straightedge

Odd proportion angles (there is a proportion of the steam angles) of the element 3 (which may be 5,7,9, ...) and the base angle $45 ^ o$ (which can be any angle which is obtainable by means of compass and straightedge)

The angle CAB $45 ^ o$ can be obtained with a compass and unmarked ruler, he added angles (each have $45 ^ o$) DAC and EAD, obtained angle EAB $135 ^ o$is the starting angle

Merge points E (D, C, B) and get a longer ED (DC, CB)

Along the DC from the point C draw is normal that intersects the segment AB, the intersection is a point G

Divider AG and from point G draw a circular arc to a longer EA and H get the point, and the arc GH, join the dots G and H and get along GH

Longer GH (ED, DC, CB) are equal, the arc EB's first circular arc can be made smaller or larger with a constant radius AB, arc GH is the second circular arc can be made smaller or larger with a constant radius AG

INCREASING THE ANGLE
the starting angle EAB add angle FAE $15 ^ o$ get the angle FAB $150 ^ o$ - continued in the next post

2. ## Re: proportioned angles

- previous post was in error -

Required accessories - pencil, compass, unmarked straightedge

basic angle - can be any angle that can be construction using compass and unmarked straightedge, angle CAB $45 ^ o$

starting angle - sum of 2, 3, 4, 5, ... basic angles , EAB $135 ^ o$
sum angles CAB $45 ^ o$ DAC $45 ^ o$ EAD $45 ^ o$

difference angle - the angle which increases or decreases the starting angle. difference starting angle and the angle of whom do not know the measure , this angle is known to see a procedure HAB $30 ^ o$

straightedge AB is divided into three parts AF , how we have a basis in the angles starting angle

divider AF from point A the circular arc FG

section straightedge AH the circular arc FG , point I

straightedges FG , ED
--------------------------------------
will continue - if there are errors

3. ## Re: proportioned angles

basic angle CAB $45^o$

starting angle EAB $135^o$ consists of the sum of the angles CAB $45^o$ DAC $45^o$EAD $45^o$

DC straightedge the normal to the point D , gets the point F

AF divider from point A, we get the point G

divider AB from point F, divider AB from point G, we get the point H

HG divider from point H, creates a circular arc FG

difference angle IAB $30^o$

section IA and longer circular arc FG is a point J

4. ## Re: proportioned angles

What was your purpose in posting this? Do you have a question? You are doing a construction but you do not say what you are trying to construct.

5. ## Re: proportioned angles

applies this photo

bisection angle DAC is obtained by point J
along AJ
GF section circular arc and along the AJ, obtained point L
AF divider, from the point J, we get the point O
divider AF, from point A circle c1
divider GL, from the point J, the circuit d1, get the points P and Q

6. ## Re: proportioned angles

So you are just going to keep chattering without telling us what you are talking about?

7. ## Re: proportioned angles

Originally Posted by HallsofIvy
So you are just going to keep chattering without telling us what you are talking about?
I found the proportion of angles to us to make sure that the process has errors