In the given figure, $$\angle B = 110^\circ$$; $$\angle C = 80^\circ$$; $$\angle F = 120^\circ$$; $$\angle ADC = 30^\circ$$ $$2\angle DGF = \angle DEF$$. The measure of $$\angle BHF$$ is

In the quadrilateral $$DEFG$$ the angle at $$D$$ is vertically opposite to $$\angle ADC$$ and so equals $$30^\circ$$, giving $$\angle DEF + \angle DGF = 360^\circ - 30^\circ - 120^\circ = 210^\circ$$. Using $$\angle DEF = 2\angle DGF$$ we get $$\angle DEF = 140^\circ$$, so the angle between $$ED$$ and the line $$HEF$$ on the other side is $$40^\circ$$. Applying the angle sum of the quadrilateral $$HBCE$$, $$\angle BHF = 360^\circ - 110^\circ - 80^\circ - 40^\circ = 130^\circ$$.