XA = YB = XY.
First, let’s construct the segment AB and choose a random point on CB:
Now, let’s draw a line through D, that is parallel to AC. We’ll label the intersection of the parallel line and the circle E.
Let’s construct another circle with radius DE, but centered at E. Notice that DE and DB are equal because they are both radii of the same circle.
We’ll label the intersection of our new circle with AB as the point F. Let’s construct a line parallel to DE that goes through the point F.
Construct a line through EF and another line parallel to EF through D. Hide the circles to make the drawing less cluttered.
Notice that EF and DG are equal because they were constructed by parallel lines. Similarly, DE and FG are equal. From before, we know that DE and EF are equal, so the figure below is a rhombus.
Construct a line through BG and label its intersection with AC as X. Construct a line parallel to GD that runs through X. Label its intersection with BC as Y.
How do we know that X and Y are the points we are looking for? From similar triangles:
Triangle GDB is similar to triangle XYB. Also, triangle GDB is an equilateral triangle because GD = DB. Therefore, triangle XYB is also an equilateral triangle and XY = YB.
Also, triangle BGF is similar to triangle BXA. Since GD = DB = GF, we have that XY = YB = AX.
Questions? E-mail me.