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Can you help me to determine the angle marked with a question mark?

$\overline {AB}$ and $\overline {DE}$ are parallel

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    what did you try?2017-01-11
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    @romanemul: Assuming $AB$ and $ED$ are parallel, draw a line through $C$ parallel to ...2017-01-11
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    and find the 2 parts of angle at $C$2017-01-11

4 Answers 4

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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 $\angle ABC+(180^\circ-\angle EDC)$.

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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?

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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 $40^\circ+60^\circ=\color{red}{100^\circ}$

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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:

$\angle{DCB}=360^{\circ}-50^{\circ}-90^{\circ}-120^{\circ}=100^{\circ}$