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 [TEX] 45 ^ o[/TEX] (which can be any angle which is obtainable by means of compass and straightedge)

The angle CAB [TEX]45 ^ o[/TEX] can be obtained with a compass and unmarked ruler, he added angles (each have [TEX]45 ^ o[/TEX]) DAC and EAD, obtained angle EAB [TEX]135 ^ o[/TEX]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 [TEX]15 ^ o[/TEX] get the angle FAB [TEX]150 ^ o[/TEX] - continued in the next post

- 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 [TEX] 45 ^ o[/TEX]

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

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 [TEX] 30 ^ o[/TEX]

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
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will continue - if there are errors

basic angle CAB [tex]45^o[/tex]

starting angle EAB [tex]135^o[/tex] consists of the sum of the angles CAB [tex]45^o[/tex] DAC [tex]45^o [/tex]EAD [tex]45^o[/tex]

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 [tex]30^o[/tex]

section IA and longer circular arc FG is a point J

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