Langley's Adventitious Angles is a puzzle in which one must infer an angle in a geometric diagram from other given angles. It was posed by Edward Mann Langley in The Mathematical Gazette in 1922.
The problem In its original form the problem was as follows:
A B C {\displaystyle ABC} is an isosceles triangle with ∠ C B A = ∠ A C B = 80 ∘ . {\displaystyle \angle {CBA}=\angle {ACB}=80^{\circ }.}
C F {\displaystyle CF} at 30 ∘ {\displaystyle 30^{\circ }} to A C {\displaystyle AC} cuts A B {\displaystyle AB} in F . {\displaystyle F.}
B E {\displaystyle BE} at 20 ∘ {\displaystyle 20^{\circ }} to A B {\displaystyle AB} cuts A C {\displaystyle AC} in E . {\displaystyle E.}
Prove ∠ B E F = 30 ∘ . {\displaystyle \angle {BEF}=30^{\circ }.}
Solution The problem of calculating angle ∠ B E F {\displaystyle \angle {BEF}} is a standard application of Hansen's resection. Such calculations can establish that ∠ B E F {\displaystyle \angle {BEF}} is within any desired precision of 30°, but being of only finite precision, always leave doubt about the exact value. A direct proof using classical geometry was developed by James Mercer in 1923. This proof involves drawing one additional line, and then making repeated use of the fact that the internal angles of a triangle add up to 180° to prove that several triangles drawn within the large triangle are all isosceles.
Draw B G {\displaystyle BG} at 20 ∘ {\displaystyle 20^{\circ }} to B C {\displaystyle BC} intersecting A C {\displaystyle AC} at G {\displaystyle G} and draw F G . {\displaystyle FG.} (See figure on the lower right.) Since ∠ B C G = 80 ∘ {\displaystyle \angle {BCG}=80^{\circ }} and ∠ C B G = 20 ∘ {\displaystyle \angle {CBG}=20^{\circ }} then ∠ B G C = 80 ∘ {\displaystyle \angle {BGC}=80^{\circ }} and triangle B C G {\displaystyle BCG} is isosceles with B C = B G . {\displaystyle BC=BG.}
Since ∠ B C F = 50 ∘ {\displaystyle \angle {BCF}=50^{\circ }} and ∠ C B F = 80 ∘ {\displaystyle \angle {CBF}=80^{\circ }} then ∠ B F C = 50 ∘ {\displaystyle \angle {BFC}=50^{\circ }} and triangle B C F {\displaystyle BCF} is isosceles with B C = B F . {\displaystyle BC=BF.}
Since ∠ F B G = 60 ∘ {\displaystyle \angle {FBG}=60^{\circ }} and B F = B G {\displaystyle BF=BG} then triangle B G F {\displaystyle BGF} is equilateral. Since ∠ B G E = 100 ∘ {\displaystyle \angle {BGE}=100^{\circ }} and ∠ G B E = 40 ∘ {\displaystyle \angle {GBE}=40^{\circ }} then ∠ G E B = 40 ∘ {\displaystyle \angle {GEB}=40^{\circ }} and triangle B G E {\displaystyle BGE} is isosceles with G B = G E . {\displaystyle GB=GE.}
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