Angle_20

Angle problem

Let ABC be an isosceles triangle (AB=AC) with angle(BAC) = 20. Point D is on side AC such that
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.

Bibliography

[CoxGrei] Coxeter, H. S. M. and Greitzer, S. L. Geometry Revisited. Washington DC, Math. Assoc. Amer., 1967
[Rike] Tom Rike An Intriguing Geometry Problem Preprint, 2002

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