A, B, C, D are collinear points in this order. Draw the regular triangles ABE and CDF on one side of the line. Let G denote the intersection of the circles ACE and BDF on the same side of the line ABCD. Prove that < AGD=120 °.
A, B, C, D are collinear points in this order. Draw the regular triangles ABE and CDF on one side of the line. Let G denote the intersection of the circles ACE and BDF on the same side of the line ABCD. Prove that < AGD=120 °.
If AB = BC = CD, and if "regular triangle" means equilateral/equiangular, then angle AGD is really 120 degrees. Points E, G and F are collinear too. Triangle BGC is a regular triangle too.