Find the intersection line between the "beta" plane and the second bisecting.
to the second bisector we call him alpha, and locate his traces on the Earth line.
We apply the general procedure of intersection of planes, we note that the points of cut of the traces coincide. Therefore, we need a second common point to the two planes, to find their line of intersection.
This second point, we will give the intersection of a line of the bisector, with the beta plane.
We draw a straight of this bisector parallel to the Earth line.
The second bisector forms 45 degrees with the projection planes, so the dimensions and the distances of all its points, coincide with the horizontal plane of projection.
We perform the intersection of the straight R, with the beta plane.
To do this, we draw a horizontal plane on this line.
We found the intersection line of this Delta plane with the beta plane.
at the cutting point of this intersection line S, with the straight R, we will obtain a point B, common to the three planes.
Uniting this point with the point of the traces v two and H one, draw the line of intersection searched.
procedure in the dihedral system.
We point out the alpha traces on the Earth line, which correspond to the traces of the second bisector.
We apply the general procedure of intersection of planes, at the point of cut of the trace Alpha one with the trace beta one, we will obtain the trace H one, of the line of intersection, and the point of cut of the vertical traces of the planes generates the trace v two.
These two traces match. Therefore, we need a second common point to the two planes, to find their line of intersection.
To do this, we draw a straight any of the second bisector, and find its point of intersection with the beta plane.
By any dimension, the vertical projection of a line parallel to the Earth's lines, contained in this bisector, is drawn.
As we have seen before, in the second bisector the dimensions and distances of all its points coincide. Therefore, the horizontal projection R one, will coincide with the vertical R two.
By this line we draw a horizontal plane.
That will generate a single vertical trace, Delta two.
We found the intersection line of the Delta and beta planes.
By having only one vertical trace, this intersection line will have its horizontal projection parallel to the beta one trace.
At the cutting point of the projection s one, with the projection R one, we will obtain B one, which is the horizontal projection of a common point to the three planes.
We draw your vertical projection, b two, over R two. Coinciding with both projections.
Unite B One with H one, to trace I one.
and Uniting B two with V two, we will draw the vertical projection of the searched intersection line.
Its two projections also coincide, and will be discontinuous because they pass through the second and fourth quadrant.