Find the angles marked with a question mark shown in Fig. 17.27

In $\triangle \mathrm{CEB}:$

$\angle \mathrm{ECB}+\angle \mathrm{CBE}+\angle \mathrm{BEC}=180^{\circ} \quad$ (angle sum property of a triangle)

$40^{\circ}+90^{\circ}+\angle \mathrm{EBC}=180^{\circ}$

$\therefore \angle \mathrm{EBC}=50^{\circ}$

Also, $\angle \mathrm{EBC}=\angle \mathrm{ADC}=50^{\circ}$ (opposite angle of a parallelogram)

In $\triangle \mathrm{FDC}:$

$\angle \mathrm{FDC}+\angle \mathrm{DCF}+\angle \mathrm{CFD}=180^{\circ}$

$50^{\circ}+90^{\circ}+\angle \mathrm{DCF}=180^{\circ}$

$\therefore \angle \mathrm{DCF}=40^{\circ}$

Now, $\angle \mathrm{BCE}+\angle \mathrm{ECF}+\angle \mathrm{FCD}+\angle F D C=180^{\circ}$ (in a parallelogram, the sum of alternate angles is $180^{\circ}$ )

$50^{\circ}+40^{\circ}+\angle \mathrm{ECF}+40^{\circ}=180^{\circ}$

$\angle \mathrm{ECF}=180^{\circ}-50^{\circ}+40^{\circ}-40^{\circ}=50^{\circ}$