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Question
Find the angles marked with a question mark shown in Fig. 17.27
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Solution
\[\text{ In } \bigtriangleup CEB: \]
\[\angle ECB + \angle CBE + \angle BEC = 180° (\text{ angle sum property of a triangle })\]
\[40° + 90° + \angle EBC = 180° \]
\[ \therefore \angle EBC = 50° \]
\[\text{ Also }, \angle EBC = \angle ADC = 50° \left( \text{ opposite angle of a parallelogram } \right)\]
\[\text{ In } \bigtriangleup FDC: \]
\[\angle FDC + \angle DCF + \angle CFD = 180° \]
\[50° + 90° + \angle DCF = 180° \]
\[ \therefore \angle DCF = 40° \]
\[\text{ Now }, \angle BCE + \angle ECF + \angle FCD + \angle FDC = 180° (\text{ in a parallelogram, the sum of alternate angles is } 180° )\]
\[50° + 40° + \angle ECF + 40° = 180° \]
\[\angle ECF = 180° - 50° + 40° - 40° = 50° \]
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