Problem 1.4.3

Here is how I thought about the problem.    There are probably other ways to get the construction.

1.   We are given an angle   and a segment equal in length to the  perimeter.   We want to construct an isosceles triangle.

2.   Think about the problem as solved and work backward.     I began with some isosceles triangle that had the given vertex angle.    Think about folding the lateral sides outward and into the  same line as the base of the triangle.

Had this been the desired triangle, that line across the bottom would be the perimeter.   But, woe is me, it is not the desired triangle so we do some more.

Well, this still is not anything we could construct.    Add a parallel line across the top through the vertex of the  isosceles triangle.

Those slant lines are beginning to look like angle bisectors.  The MUST be because the obtuse triangle on each side is isosceles (we constructed it by folding out the sides.   And there are alternate interior angles of parallel lines.        Let's see,   Add lines through each of those external base points to complete a rhombus on each side (I am not sure we need them.)

Let's see,  if we started just with the given angle, we could construct its angle bisector and then construct a perpendicular line to the angle bisector through the vertex of the angle

Now construct the angle bisectors for the angles on each side

Now the task is   to 'fit'  the  segment representing the perimeter across to connect to the two green angle bisectors.    That segment will have the bisector of your give angle as a perpendicular bisector.    Take your given perimeter line segment, find its midpoint, and construct a copy of the segment somewhere with its midpoint on the angle bisector of the given angle.

Now, we have the tools from parallel projection to copy this segment up to where its endpoints are on the green angle bisectors:

Does that get it?