

A difficult geometry problem 


Request
Below I publish solutions of a geometry problem.
You will find analytical (numeric), trigometrical but no geometrical solutions.
In case you have a new approach to this problem please let me know.
Your work will be published here.
Thanks in advance!
Problem
Above is pictured an equilateral triangle ABC and square ADEF.
LBFC=45^{o}.
Proof that LECF=15^{o} deg.
How to construct this figure?
 point G is the center of AB.
 extend edge BA.
 rotate GC 90^{o} counterclockwise.
 HG=GC.
 construct circle through points H, B, C.
 F on circle such, that LHAF=30^{o} so, AF perpendicular to AC.
 rotate AF clockwise, intersection with AC is point D.
 rotate AF around centers F and D, intersection is point E.
LCHB = LCFB so are on circle.
Let AF=2 and AB=2p.
Given LBFC=45^{o} enables calculation of p.
Knowing p, the coordinates of points C,E,F can be calculated.
Then Lx (LECF) may be calculated.
For these calculations using complex numbers is convenient.
Calculation of p
Point A is designated as (0,0).
Coordinates of other points follow easily.
Multiply EB by (1+i) which caused 45 degrees ratation.
The product vector has the same argument as vector EC.
Lx = LECD is the difference of vectors CE and CD.
Lx=LDCE=15^{o}
Request for help
Problem posted in Facebook group "Classical Mathematics" of Omid Motahed.
Below I publish the responses.
Response (1)
Dejan Piscevic.
He applies the sine rule in three triangles to obtain 3 equations with unknowns x, b and z.
Presents solution x=15^{0} but no calculations are included.
I continue
See figure above for the calculation of b.
Unclear is yet if Lx exactly is 15^{0}.
Not rounding the roots in the formulas for b and tan(x) may answer this question.
Construct equilateral triangle AMD, M on AB.
Draw circle with center M through points E and F.
Next, prove that point C is on this circle:
LECF spans arc EF so Lx = 30/2 = 15^{0}.
Response (2)
Wong Sii Hiang:
Response (3)
Hans Karl Abele:
Response (4)
Biro Istvan :
Response (5)
Viktor Martoni
Response (6)
Joszef Nagy Vonnak

