This solution has been reported in [Leikin].
The proof is practically by construction.
Starting from BC, form a sequence of isosceles triangles, BCF, CFD, FDE, DEA. Is it possible? Let us check the angles.
For ΔBCF to be isosceles, suffice it to pick ∠BCF = 20°, for then ∠BFC = 80° = ∠CBF.
If point D on AC is such that DF = FC (and such point exists!), then, since ∠DCF = 60°, ΔCDF is equilateral, with all the angles 60° and sides equal.
If E' on AB (not shown) is such that DF = DE' (and such a point exists), then ΔFDE' is isosceles, ∠DFE' = 40° = ∠FE'D and ∠FDE' = 100°.
It follows that ∠ADE' = 20° = ∠AE'D. Therefore, ΔADE' is isosceles and DE' = AE'. But, by construction, DE' = DF = CF = BC. So, indeed, E = E'.
Since ΔCFD is equilateral, CD = BC = DE, so that ΔCDE is also equilateral with ∠CDE = 100° + 60° = 160° which makes ∠CED = 10°. Now, ∠AEC = 140° + 10° = 150°.
Reference
- R. Leikin, Dividable Triangles - What Are They?, Mathematics Teacher, May 2001, pp. 392-398.
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