This solution has been reported in [Which Way Did the Bicycle Go?].
Add a couple of triangles ACU and AUV equal to ΔABC and mark point F on AV, AF = AE. ∠EAF = 60° which makes ΔEAF equilateral. Thus, EF = AE = BC = CU. Also, because of the symmetry, AF||CU so that CEFU is a parallelogram. And again, with a nod to the symmetry, the angles CEF and UFE are right, and so are the angles ECU and FUC. Importantly, ∠ECU = 90°. This solves the problem:
∠ACE = ∠ECU - ∠ACU = 90° - 80° = 10°,
∠AEC = 180° - 20° - 10° = 150°.
Reference
- J. Konhauser, D. Velleman, S. Wagon, Which Way Did the Bicycle Go?, MAA, 1996, #15
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