My May diary included a brainteaser based on the figure at left.
Given that the two shaded triangles are equilateral, what is the shaded angle? No, you can't use a protractor.
Life is easier if we label some points here.
Now: This is one of those cases where there's a heuristic solution and a properly rigorous, Euclidean one.
Heuristic solution. Since we are not given any information about the relative dimensions of the two equilateral triangles, presumably the answer is the same whatever those dimensions are.
So shrink the right-hand triangle down to microscopic size. Now CG is CE, near enough; FE is FD, near enough; and the angle FHG is the angle FDE, near enough; and that is 120° (180° minus the 60° at CDF).
Rigorous solution. First, note that the triangles CGD and FED are congruent. Proof: Their two sides GD, ED are equal; so are their other two sides DF and DC. Since angle CDG and angle FDE are both 120°, the triangles are congruent on the side-angle-side principle.
From this congruence it follows that angle CGD and angle FED are equal.
From that it follows that the triangles CGD and CEH are similar. Not congruent, just similar: One pair of angles (DCG, HCE) are the same angle, and we've just shown that another pair are equal.
So angles CDG, CHE are equal.
But angle CDG is 120°. So angle CHE must be 120°, too; and therefore so must angle FHG.