The figure described has 6 equal length segments.
Draw segments CE and EF which are bases of isosceles triangles EBC and EAF.
angle EBC = 90-60 = 30 so angle BEC=75
angle EAF = 90+60-60 = 90 so angle AEF = 45
angle CEF = angles BEC + angle BEA + angle AEF = 75+60+45 = 180
Since the angle is 180 degrees, the points C, E and F are collinear.
Posted by Jer
on 2013-10-24 14:19:49