Note: Figure not drawn to scale. In triangle , bisects with on . Point lies on line with between and . The measure of is , and the measure of is . If is the measure of , what is the value of ?

Solution available with step-by-step explanation

SAT Lines, Angles & Triangles Question 12 | Free SAT Prep | Aniko

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Question 12·Hard·Lines, Angles, and Triangles

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Note: Figure not drawn to scale.

In triangle $ABC$ , $AD$ bisects $\angle BAC$ with $D$ on $\overline{BC}$ . Point $E$ lies on line $BC$ with $C$ between $B$ and $E$ . The measure of $\angle ABD$ is $31^\circ$ , and the measure of $\angle ACE$ is $123^\circ$ . If $x$ is the measure of $\angle ADB$ , what is the value of $x$ ?

Your answer

(Express the answer as an integer)

For angle-chasing geometry questions, start by carefully marking all given angles and key facts (like angle bisectors and straight lines) on a quick sketch, even if the figure is not to scale. Use basic relationships systematically: angles on a straight line sum to $180^\circ$ , the angles in a triangle sum to $180^\circ$ , and an exterior angle of a triangle equals the sum of the two non-adjacent interior angles. Pay close attention to which point is the vertex of the angle you’re asked to find, and whether the angle is interior or exterior to the triangle, to avoid stopping one step too early with the wrong angle.

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Connect ∠ABD to triangle ABC

Since $D$ lies on $\overline{BC}$ , how does $\angle ABD$ compare to $\angle ABC$ in triangle $ABC$ ?

Use the straight line at C

Because $B$ , $C$ , and $E$ lie on the same line, rays $CB$ and $CE$ form a straight line. How are $\angle ACB$ and $\angle ACE$ related when they share side $CA$ ?

Find the angle at A and then use the bisector

Once you know $\angle ABC$ and $\angle ACB$ , use the triangle angle sum to find $\angle BAC$ . Then use the fact that $AD$ bisects $\angle BAC$ to find $\angle BAD$ and $\angle DAC$ .

Think about an exterior angle at D

In triangle $ADC$ , extend side $CD$ through $D$ to $B$ . The angle at $D$ formed by $DA$ and $DB$ is an exterior angle. How does this angle relate to the two non-adjacent interior angles of triangle $ADC$ ?

Desmos Guide

Compute the angle at C

In Desmos, type c = 180 - 123 to represent $\angle ACB$ as $c$ (in degrees).

Find the angle at A and its half

Type a = 180 - 31 - c to represent $\angle BAC$ as $a$ . Then type h = a/2 to represent each half of $\angle BAC$ (this is $\angle BAD$ and $\angle DAC$ ).

Express x in terms of the triangle’s angles

Use the exterior angle relationship in triangle $ADC$ by typing x = h + c. The value Desmos shows for x is the measure of $\angle ADB$ in degrees.

Step-by-step Explanation

Relate the given angles at B and C to triangle ABC

Because $D$ lies on segment $BC$ , ray $BD$ is the same as ray $BC$ , so

\angle ABD = \angle ABC = 31^\circ.

At $C$ , points $B$ , $C$ , and $E$ are collinear with $C$ between $B$ and $E$ , so rays $CB$ and $CE$ form a straight line. Therefore, $\angle ACB$ and $\angle ACE$ are supplementary (add to $180^\circ$ ):

\angle ACB = 180^\circ - 123^\circ = 57^\circ.

Find the angle at A in triangle ABC

Now use the fact that the interior angles of a triangle add to $180^\circ$ :

\angle BAC + \angle ABC + \angle ACB = 180^\circ.

Substitute $\angle ABC = 31^\circ$ and $\angle ACB = 57^\circ$ :

\angle BAC + 31^\circ + 57^\circ = 180^\circ \\[0.7em] \angle BAC + 88^\circ = 180^\circ \\[0.7em] \angle BAC = 92^\circ.

Use the angle bisector to split ∠BAC

Segment $AD$ bisects $\angle BAC$ , so it divides $\angle BAC$ into two equal angles:

\angle BAD = \angle DAC = \frac{\angle BAC}{2} = \frac{92^\circ}{2} = 46^\circ.

Also note that $\angle ACD$ is the same as $\angle ACB$ (ray $CD$ lies along $CB$ ), so $\angle ACD = 57^\circ$ .

Apply the exterior angle theorem at D to find ∠ADB

Look at triangle $ADC$ . At vertex $D$ , side $CD$ is extended along the line to point $B$ , so $\angle ADB$ is an exterior angle of triangle $ADC$ .

For any triangle, an exterior angle equals the sum of the two interior angles that are not adjacent to it (the two angles at the other vertices). Here, those are $\angle DAC$ and $\angle ACD$ :

\angle ADB = \angle DAC + \angle ACD.

Substitute the values you found:

\angle ADB = 46^\circ + 57^\circ = 103^\circ.

So $x = 103$ .

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Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation