## May 2020

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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, youcan'tuse a protractor.

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**• Solution**

Life is easier if we label some points here.

That's better.

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).

Answer: 120°

**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.

Answer: 120°