Edward Mann Langley, founder of the *Mathematical Gazette*, posed this problem in its pages in 1922:

ABCis an isosceles triangle.B=C= 80 degrees.CFat 30 degrees toACcutsABinF.BEat 20 degrees toABcutsACinE. Prove angleBEF= 30 degrees.

(Langley’s description makes no mention of *D*; perhaps this is at the intersection of *BE* and *CF*.)

A number of solutions appeared. One, offered by J.W. Mercer in 1923, proposes drawing *BG* at 20 degrees to *BC*, cutting *CA* in *G*. Now angle *GBF* is 60 degrees, and angles *BGC* and *BCG* are both 80 degrees, so *BC* = *BG*. Also, angles *BCF* and *BFC* are both 50 degrees, so *BF* = *BG* and triangle *BFG* is equilateral. But angles *GBE* and *BEG* are both 40 degrees, so *BG* = *GE* = *GF*. And angle *FGE* is 40 degrees, so *GEF* is 70 degrees and *BEF* is 30 degrees.