Question 42·Hard·Lines, Angles, and Triangles

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In triangle $ABC$ , point $D$ lies on segment $\overline{AB}$ and point $E$ lies on segment $\overline{AC}$ . Segment $\overline{DE}$ is drawn.

The measure of angle $BAC$ is $40^\circ$ . The measure of angle $ABC$ is $(3x-5)^\circ$ . The measure of angle $AED$ is $(2x+10)^\circ$ .

Which choice gives the value of $x$ for which $\overline{DE}$ is parallel to $\overline{BC}$ ?

Note: Figure not drawn to scale.

When a segment is parallel to a side of a triangle, look for matching angles created by a transversal (here, line $AC$ ). Convert the needed angle at $C$ into an expression using the triangle angle sum, then set it equal to the given angle at the intersection point (here, $\angle AED$ ) and solve.

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Identify the angle at $C$ that matches $\angle AED$

If $\overline{DE}$ is parallel to $\overline{BC}$ , then $AC$ intersects both parallel lines. Which angle at $C$ uses side $AC$ ?

Write the third angle of triangle $ABC$ in terms of $x$

Use $\angle A + \angle B + \angle C = 180^\circ$ with $\angle A = 40^\circ$ and $\angle B = (3x-5)^\circ$ .

Set the matching angles equal

When the lines are parallel, the angle at $E$ should equal the angle at $C$ you found. Create an equation and solve for $x$ .

Desmos Guide

Enter the two expressions to be set equal

In Desmos, graph the two lines:

$y=2x+10$
$y=145-3x$

Find the intersection

Click the intersection point of the two graphs. The $x$ -coordinate of the intersection is the value of $x$ that makes the angles equal.

Connect the intersection to the geometry condition

Use that $x$ -value because $\angle AED = \angle ACB$ is the condition that occurs when $\overline{DE} \parallel \overline{BC}$ (corresponding angles with transversal $AC$ ).

Step-by-step Explanation

Find $\angle ACB$ in terms of $x$

In triangle $ABC$ , the angles sum to $180^\circ$ :

\angle ACB=180^\circ-\angle BAC-\angle ABC

Substitute the given expressions:

\angle ACB=180-40-(3x-5)=145-3x

Use the parallel-lines angle relationship

If $\overline{DE} \parallel \overline{BC}$ , then line $AC$ acts as a transversal of the parallel lines $DE$ and $BC$ .

So the angle formed by $AC$ and $DE$ at $E$ equals the angle formed by $AC$ and $BC$ at $C$ :

\angle AED = \angle ACB

Set up and solve the equation

Set the angle measures equal:

2x+10 = 145-3x

Solve:

5x=135 \quad \Rightarrow \quad x=27

Therefore, the correct choice is $\boxed{27}$ .

Question 42·Hard·Lines, Angles, and Triangles

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation