For reference: The figure, ABCD is a parallelogram. T and C are points of tangency. Calculate x.
My progress:
I marked the angles I could find…but the main one is missing
$\triangle AEC (isosceles) \implies \measuredangle CAE = 80^\circ~and~ \measuredangle TAC = 100^\circ\\
TAJC (cyclic quadrilateral) \implies \measuredangle CJT = 80^\circ~and~\measuredangle GJC =100^\circ\\
\therefore JBGI: \measuredangle G = \measuredangle B = 80^\circ~and \measuredangle I = 100^\circ\\
\measuredangle MCE = \frac{\overset{\LARGE{\frown}}{CE}}{2} =40^\circ\\
\measuredangle CEH = 130^\circ$
Best Answer
Using your diagram:
Let's prove that triangle CJI is similar to triangle ACH. HT is perpendicular to JG, but JG is parallel to CI, so that HT and CI are perpendicular. Let those two lines intersect at a point X. Thus, angle IXH = 90 and so is angle IKH, so that IKXH is cyclic and so angle AHC = angle XHK = angle XIK = angle CIJ. Then, angle CJI = angle CJK = 90 - angle JCK = 90 - angle HCM = angle ACH. Thus, they're similar, and we conclude that CJ/AC = CI/AH, ie. that TJ / TA = JG / AH. This implies that triangles TJA and TGH are similar and that JA and GH are parallel - so that angle x is equal to angle AJC which is half of 80 which is 40.