Question

in the diagram below, $overline{ab} parallel overline{cd}$, $overline{ad} parallel overline{bc}$, $overline{ed}$ bisects $angle adc$, $overline{ec}$ bisects $angle dcb$ and $mangle a = 66^circ$. find $mangle dec$.
you may assume lines that appear straight are straight, but the figure is not otherwise drawn to scale.

Explanation:

Step1: Find $\angle ADC$

In parallelogram $ABCD$, consecutive angles are supplementary:
$$m\angle A + m\angle ADC = 180^\circ$$
$$66^\circ + m\angle ADC = 180^\circ$$
$$m\angle ADC = 180^\circ - 66^\circ = 114^\circ$$

Step2: Find $\angle EDC$

$\overline{ED}$ bisects $\angle ADC$, so:
$$m\angle EDC = \frac{1}{2}m\angle ADC$$
$$m\angle EDC = \frac{1}{2} \times 114^\circ = 57^\circ$$

Step3: Find $\angle DCB$

In parallelogram $ABCD$, opposite angles are equal:
$$m\angle DCB = m\angle A = 66^\circ$$

Step4: Find $\angle ECD$

$\overline{EC}$ bisects $\angle DCB$, so:
$$m\angle ECD = \frac{1}{2}m\angle DCB$$
$$m\angle ECD = \frac{1}{2} \times 66^\circ = 33^\circ$$

Step5: Find $\angle DEC$

Sum of angles in $\triangle DEC$ is $180^\circ$:
$$m\angle DEC + m\angle EDC + m\angle ECD = 180^\circ$$
$$m\angle DEC + 57^\circ + 33^\circ = 180^\circ$$
$$m\angle DEC = 180^\circ - 90^\circ = 90^\circ$$

Answer:

$m\angle DEC = 90^\circ$