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I am trying to solve the following problem.
Given: $$\overleftrightarrow{AB} \parallel \overleftrightarrow{DE}$$ And the measures of angles $$\angle BAC = 42 ^{\circ}$$ $$\angle EDC= 54 ^{\circ}$$ Find the measure of angle $$\angle ACD$$ How would you approach to solve the problem. I have tried as recommended on book to assume that there is one parallel line passing through point $C$, but i couldn't produce an answer. What is your advice?
Attached follows the geometric representationFigure 1 Geometric Model

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  • $\begingroup$ Draw a line through $C$ which is parallel to the other lines. Then the angle you want is the sum of 2 angles whose measure are easy to find. $\endgroup$
    – Paul
    Aug 18, 2018 at 12:38
  • $\begingroup$ @Paul I appreciate. $\endgroup$
    – JFC
    Aug 18, 2018 at 12:54

1 Answer 1

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There is a line perpendicular to $AB$ and $DE$ passing through $C$. Say this line meets $AB$ at $X$ and $DE$ at $Y$. Then using the two right-angled triangles, $\angle ACX=90-42=48^{\circ}$ and $\angle DCY=90-54=36^{\circ}$. Since angles on a straight line sum to $180^{\circ}$, $\angle ACD=180-48-36=96^{\circ}$.

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  • $\begingroup$ On the same figure if we make $\overleftrightarrow{AC}$ Transversal and draw parallel line $l$ through point $C$, this seems that angle between Transversal $\overleftrightarrow{AC}$ and parallel line $l$ is congruent to $\angle BAC$ $\endgroup$
    – JFC
    Aug 18, 2018 at 13:04
  • $\begingroup$ i mean inner right angle between Transversal $\overleftrightarrow{AC}$ and parallel line $l$ $\endgroup$
    – JFC
    Aug 18, 2018 at 13:06
  • $\begingroup$ @JFC Yes, you're right, because alternate angles are equal $\endgroup$
    – A. Goodier
    Aug 18, 2018 at 13:19
  • $\begingroup$ Thanks for the answer. $\endgroup$
    – JFC
    Aug 18, 2018 at 13:55

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