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Langley's adventitious angles
A seemingly simple problem first posed in 1922 by E. M. Langley in connection with an isosceles triangle. In its original form, it is stated as follows: ABC is an isosceles triangle. B = C = 80°. CF at 30° to AC cuts AB in F. BE at 20° to AB cuts AC in E. Prove angle BEF = 30°. (No mention is made of D. Perhaps it is at the intersection of BE and CF.)
A number of solutions appeared shortly after, including this one given by J. W. Mercer: Draw BG at 20° to BC, cutting CA in G. Then angle GBF = 60° and angles BGC and BCG are 80°. So BC = BG. Also angle BCF = angle BFC = 50°, so BF = BG and triangle BFG is equilateral. But angle GBE = 40° = angle BEG, so BG = GE = GF. And angle ° = 40°, hence GEF = 70° and BEF = 30°.
Reference
- Langley, E. M. "Problem 644." Mathematical Gazette, 11: 173, 1922.
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• MATHEMATICS
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