Quadrilateral with given angles We are looking for angles x and y.
I have found the values of the following angles:
BEA = 74,
BDA = 64,
ACD = 68,
ECD = 112,
plus the relationship $x+y = 68$.
All other angles equations, from triangles or the sum of angles in the quadrilateral (360) end up in the same equation!
I have found through Geogebra that $x=18$ and $y=50$ but I can't figure out a second relationship to determine them geometrically!
Does anyone have any ideas?
Thank you!
If you are looking for the proof by elementary geometry, please see Hiroshi Saito(斉藤浩)'s great work about generalized Langley's problem
( https://www.gensu.co.jp/saito/challenge/pdf/3circumcenter_d20180609.pdf ). He introduced the amazing skill named &quot;3 circumcenter method&quot; invented by Ms. aerile_re(pen name), and you can find the solution of this problem in the article (Q1).
Sinx / Sin(x + 64) = Sin38 / Sin84 *Sin22 / Sin48
Multiply both denom and nom by 4Sin82
Sinx /Sin(x + 64) = 4Sin22Sin38Sin82 / (Cos6Sin48Sin82)
(4Sin22Sin38Sin82 = Sin66 REPLACEMENT)
Sinx / Sin(x + 64) = Cos24 / (4Sin48Cos6Sin82)
Sinx / Sin(x + 64) = 1 / (8Sin24Cos6Sin82)
(2Sin24*Cos6 = Sin54 REPLACRMENT)
Sinx / Sin(x + 64) = 1 / (4Sin54*Sin82)
(1/4Sin54 = Sin18 REPLACEMENT)
Sinx / Sin(x + 64) = Sin18 / Sin82
X = 18
Y = 50