Langley’s Adventitious Angles Problem

We can draw BG at 20° to BC meeting CE in G. Now it's very easy to prove that GB=GF=GE which makes <BGF=60°. In other words, G is the circum-centre of triangle BFE. BF subtends angle of 60° at G and therefore <BEF= $\frac{1}{2}$ <BGF =30°.

1. Calculate some known angles:

   ACB = 180-(10+70)-(60+20) = 20° AEB = 180-60-(50+30) = 40°

2. Draw a line from point E parallel to AB, labeling the intersection with AC as a new point F and conclude:

   FCE ACB CEF = CBA = 50+30 = 80° FEB = 180-80 = 100° AEF = 100-40 = 60° CFE = CAB = 60+20 = 80° EFA = 180-80 = 100°

3. Draw a line FB labeling the intersection with AE as a new point G and conclude:

   AFE BEF AFB = BEA = 40° BFE = AEF = 60° FGE = 180-60-60 = 60° = AGB. ABG = 180-60-60 = 60°

4. Draw a line DG. Since AD=AB (leg of isosceles) and AG=AB (leg of equilateral), conclude:

   AD = AG. DAG is isosceles ADG = AGD = (180-20)/2 = 80°

5. Since DGF = 180-80-60 = 40°, conclude:

   FDG (with two 40° angles) is isosceles, so DF = DG

6. With EF = EG (legs of equilateral) and DE = DE (same line segment) conclude:

   DEF DEG by side-side-side rule DEF = DEG = x FEG = 60 = x+x

Answer X = 30 degree .