determine the size of angle 
Can you help me to determine the angle marked with a question mark?
$\overline {AB}$ and $\overline {DE}$ are parallel
 A: HINT: Draw a line through $C$ that is parallel to $AB$ and $DE$. This splits the angle in question into two parts. Can you find the measures of the two parts using alternate interior angles? 
A: 
Divide $\angle C$ into angles $C_1$ and $C_2$ as shown. 


*

*Since $\overline {AB}$ and $\overline {CF}$ are parallel, then $\angle C_2=40^\circ$ by alternate interior angles. 

*You can see that the supplement of $\angle D$ is $60^\circ$. Since $\overline {DE}$ and $\overline {CF}$ are parallel, then $\angle C_1=60^\circ$ by alternate interior angles. 


Add up $\angle C_1$ and $\angle C_2$ and you get $40^\circ+60^\circ=\color{red}{100^\circ}$
A: Assuming lines $AB$ and $DE$ are parallel, you can draw a line perpendicular to the line $DE$ (let's call this point $M$) to the point $B$ to obtain the quadrilateral $BCDM$.
Now, use the fact that the sum of the interior angles in a quadrilateral is $360^{\circ}$.
From this, you can deduce that angles $\angle{CBM}=90^{\circ}-40^{\circ}=50^{\circ}$ and $\angle{BMD}=90^{\circ}$.
Thus, the angle of the question mark is:
$\angle{DCB}=360^{\circ}-50^{\circ}-90^{\circ}-120^{\circ}=100^{\circ}$
A: Draw a line perpendicular to $AB$ through $C$.
Then $\angle A'CB=180^\circ-90^\circ-40^\circ=90^\circ-40^\circ=50^\circ$.
Do the same for $DE$ to get $\angle D'CE=90^\circ-60^\circ=30^\circ$.
Now $\angle BCD=180^\circ-(50^\circ+30^\circ)=100^\circ$.
This, conveniently, is $\angle ABC+(180^\circ-\angle EDC)$. 
