**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.

Add some more pieces:

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?