angle(DBC) = 60. Point E is on side AB such that angle(ECB) = 50. Find, with proof, the measure of

angle(EDB) [CoxGrei, p. 26].

There are several equilaterals hidden in the figure. In particular drawing parallel DH to BC we

get two symmetric on AF equilaterals. An other one, EBG=EFC, is constructed by considering the

circle at B with radius BC. By the symmetry angle(EHF) = angle(FDC)=180-(60+80)=40.

angle(EFH) = 180- angle(EFC) = 40. Thus HEFD has two pairs of adjacent sides equal, consequently

its diagonals intersect orthogonally. Thus FD bisects angle(HDF) which is 60 degrees.

For an interesting discussion and the history of the problem see [Rike] in the references below.

[Rike] Tom Rike

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