Answer 1

\[\text{ In } \bigtriangleup CEB: \]

\[\angle ECB + \angle CBE + \angle BEC = 180° (\text{ angle sum property of a triangle })\]

\[40° + 90° + \angle EBC = 180° \]

\[ \therefore \angle EBC = 50° \]

\[\text{ Also }, \angle EBC = \angle ADC = 50° \left( \text{ opposite angle of a parallelogram } \right)\]

\[\text{ In } \bigtriangleup FDC: \]

\[\angle FDC + \angle DCF + \angle CFD = 180° \]

\[50° + 90° + \angle DCF = 180° \]

\[ \therefore \angle DCF = 40° \]

\[\text{ Now }, \angle BCE + \angle ECF + \angle FCD + \angle FDC = 180° (\text{ in a parallelogram, the sum of alternate angles is } 180° )\]

\[50° + 40° + \angle ECF + 40° = 180° \]

\[\angle ECF = 180° - 50° + 40° - 40° = 50° \]