A geometry problem by Guri Singh

$"x"\quad is\quad the\quad angle\quad we\quad are\quad looking\quad for,\quad let\quad be\quad \angle DEC=x\quad \\ with\quad E\quad and\quad D\quad points\quad in\quad AB\quad and\quad AC\quad respectively,\quad \\ \angle FCB=20\quad with\quad point\quad F\quad in\quad AB.\\ Because\quad \angle CAB=20\quad and\quad AC=AB,\quad \angle ACB=\angle ABC=80\\ \angle CFB=180-\angle FCB-\angle FBC=80\Rightarrow \triangle FCB\quad isosceles\quad with\quad \\ CF=CB,\quad \angle BDC=180-\angle ACB-\angle DBC=50\Rightarrow \triangle DCB\quad isosceles\quad \\ with\quad CD=CB.\quad Hence\quad CD=CF\Rightarrow \quad \angle CDF=\angle DFC\quad and\quad \\ because\quad \angle DCF=60,\quad \triangle CDF\quad equilateral\quad with\quad DF=CD=CF.\quad \\ In\quad \triangle BCE,\quad \angle CEB=40,\quad then\quad \triangle CEF\quad isosceles\quad with\quad EF=CF,\quad \\ but\quad CF=DF,\quad then\quad \triangle DEF\quad isosceles\quad with\quad \angle DEF=\angle EDF=70\quad \\ because\quad \angle EFD=40\quad (for\quad this\quad \angle EFD=180-\angle DFB).\\ Hence\quad x=\angle DEF-CEB=70-40=\boxed { 30 } .$

That's the exact way to solve it. Actually, it was my question posted by my friend and classmate Guri Singh. Well Done Enoc Cetina