## Langley's adventitious anglesA 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.
## Related category• GEOMETRY | |||||

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