This problem is a model created to solve the problem posed in the lesson Collapsible.
In the figure CF = FB = FE. If C is moved along CB, describe the paths of F and E. Explain or prove that they are so.
This problem can be explored using GeoGebra applet. Click this link to explore before you read on.
One way to prove that FC is a straight line and perpendicular to AC is to show that FC is a part of a right triangle. To do this to let x be the measure of FCB. Because FCB is an isosceles triangle, FBC and CFB is (180-2x). This implies that EFB is 180-2x being supplementary to CFB thus CFB must be 2x. Triangle EFB is an isosceles triangle so FBC must be (180-2x)/2. Adding CFB and FBC we have x+ (180-2x)/2 which simplifies to 90. Thus, EB is perpendicular to CB.
The path of F of course is circular with FB as radius.