AMC 10 · 2008 · #24

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

isosceles-triangleequilateral-triangleangle-sum-triangle identify-subproblems ↑ Prerequisites: isosceles-triangle

📏 Long solution 💡 4 insights

Problem

Quadrilateral $ABCD$ has $AB = BC = CD$ , $m\angle ABC = 70^\circ$ and $m\angle BCD = 170^\circ$ . What is the degree measure of $\angle BAD$ ?

Pick an answer.

(A)

$\ 75$

(B)

$\ 80$

(C)

$\ 85$

(D)

$\ 90$

(E)

$\ 95$

AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: Quadrilateral ABCD has three equal consecutive sides AB = BC = CD. The corner at B measures 70 degrees and the corner at C measures 170 degrees. Find the measure of angle BAD, the corner at A.

Givens: AB = BC = CD (three consecutive sides are equal); angle ABC = 70 degrees; angle BCD = 170 degrees

Unknowns: The degree measure of angle BAD

Understand

Givens: AB = BC = CD (three consecutive sides are equal); angle ABC = 70 degrees; angle BCD = 170 degrees

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #17 Visualize Spatial Relationships

Three sides share the same length, which begs for an auxiliary construction that recycles that length. Building an equilateral triangle on CD manufactures a fourth segment of the same length plus a clean 60 degree angle. That 60 degree slice tames the 170 degree corner into 110 degrees, and the figure splits into two friendly pieces: a parallelogram and an isosceles triangle. Then angle BAD is just one corner minus a smaller one.

Execute — Answer: C

#1 Draw a Diagram 7.G.A.2 Step 1

Build an equilateral triangle on CD

Let the common length be 1, so AB = BC = CD = 1.
On side CD, construct equilateral triangle CDE with E on the same side of CD as B.
Then CE and DE also equal 1, and every angle of triangle CDE is 60 degrees.
This hands us a new segment of the shared length and a known 60 degree angle to work with.

$$CE = DE = CD = 1,\quad \angle DCE=\angle CDE=\angle DEC=60^\circ$$

💡 An equilateral triangle is a length-and-angle factory: it copies the side length and stamps out exact 60 degree angles.

#7 Identify Subproblems 7.G.B.5 Step 2

Slice the 170 degree corner

Ray CE lies inside the 170 degree corner at C, cutting off the 60 degree piece DCE.
What remains between CB and CE is angle BCE = 170 - 60 = 110 degrees.
Notice too that CE = CB = 1, so C is the apex of an isosceles pair.

$$\angle BCE=\angle BCD-\angle DCE=170^\circ-60^\circ=110^\circ,\quad CE=CB=1$$

💡 Peeling a known 60 degree wedge off the big corner leaves a plain 110 degrees behind.

#1 Draw a Diagram 8.G.A.5 Step 3

Spot a pair of parallel sides

Look at line BC as a transversal crossing line BA at B and line CE at C.
The two same-side interior angles are angle ABC = 70 degrees and angle BCE = 110 degrees, and 70 + 110 = 180.
When same-side interior angles fill a straight 180 degrees, the two lines are parallel, so BA is parallel to CE.
Since we also have AB = CE = 1, quadrilateral ABCE is a parallelogram.

$$\angle ABC+\angle BCE=70^\circ+110^\circ=180^\circ\ \Rightarrow\ BA\parallel CE,\quad AB=CE$$

💡 If the two angles hugging one side of a crossing line add to a straight angle, the lines never meet.

#7 Identify Subproblems 8.G.A.5 Step 4

Cash in the parallelogram

In parallelogram ABCE, opposite sides are equal, so EA = BC = 1.
Opposite angles are equal as well, so angle EAB = angle BCE = 110 degrees and angle AEC = angle ABC = 70 degrees.
Now EA is a fourth segment of length 1, ready to pair with the equilateral side ED.

$$EA=BC=1,\quad \angle EAB=\angle BCE=110^\circ,\quad \angle AEC=\angle ABC=70^\circ$$

💡 A parallelogram mirrors its opposite corners and its opposite sides, so facts on one side copy straight across.

#7 Identify Subproblems 8.G.A.5 Step 5

Read the isosceles triangle AED

Triangle AED has AE = 1 (from the parallelogram) and ED = 1 (the equilateral side), so it is isosceles.
Its apex angle at E is angle AED = angle AEC + angle CED = 70 + 60 = 130 degrees, since ray EC sits between EA and ED.
The two equal base angles then split the remaining 50 degrees, giving angle DAE = (180 - 130)/2 = 25 degrees.

$$AE=ED=1,\quad \angle AED=\angle AEC+\angle CED=70^\circ+60^\circ=130^\circ,\quad \angle DAE=\tfrac{180^\circ-130^\circ}{2}=25^\circ$$

💡 Equal legs force equal base angles, so once you know the top angle the two bottom angles are fixed.

#7 Identify Subproblems 7.G.B.5 Step 6

Subtract to reach angle BAD

At vertex A, ray AD lies inside the parallelogram corner EAB, so angle BAD is the big corner minus the small triangle corner: angle BAD = angle EAB - angle DAE = 110 - 25 = 85 degrees.
That matches answer choice (C).

$$\angle BAD=\angle EAB-\angle DAE=110^\circ-25^\circ=85^\circ$$

💡 The angle you want is what is left after trimming the little triangle's corner off the parallelogram's corner.

[1] #1 7.G.A.2 Let the common length be 1, so AB = BC = CD = 1. On side CD, construct equilater

[2] #7 7.G.B.5 Ray CE lies inside the 170 degree corner at C, cutting off the 60 degree piece D

[3] #1 8.G.A.5 Look at line BC as a transversal crossing line BA at B and line CE at C. The two

[4] #7 8.G.A.5 In parallelogram ABCE, opposite sides are equal, so EA = BC = 1. Opposite angles

[5] #7 8.G.A.5 Triangle AED has AE = 1 (from the parallelogram) and ED = 1 (the equilateral sid

[6] #7 7.G.B.5 At vertex A, ray AD lies inside the parallelogram corner EAB, so angle BAD is th

Review

Reasonableness: Check the four corners of ABCD. We found angle BAD = 85 degrees. The corner at D is angle ADC = angle CDE - angle ADE = 60 - 25 = 35 degrees. Then 85 + 70 + 170 + 35 = 360 degrees, exactly the angle sum a quadrilateral must have, so the answer is consistent. It is also comfortably below 120 degrees, the total the two unknown corners at A and D must share (360 - 70 - 170), which rules out any value that would leave too little for D.

Alternative: Drop the picture and use coordinates. Put B = (0,0) and C = (1,0). Then A = (cos 70, sin 70) and D = (1 + cos 10, sin 10). The vectors from A to B and from A to D give cos(angle BAD) = (AB . AD)/(|AB||AD|) = 0.0872, so angle BAD = 85 degrees. A trigonometric route also works: diagonal BD makes triangle BCD isosceles with base angles 5 degrees, so angle ABD = 65 degrees, and the Law of Cosines in triangle ABD lands on 85 degrees as well.

CCSS standards used (min grade 8)

7.G.A.2 Draw geometric shapes with given conditions, focusing on constructing triangles (Constructing the auxiliary equilateral triangle CDE on side CD to create a fourth equal length and a 60 degree angle)
7.G.B.5 Use facts about supplementary, complementary, vertical, and adjacent angles to solve for an unknown angle (Splitting the 170 degree corner into 110 degrees, and subtracting 25 degrees from 110 degrees to get angle BAD)
8.G.A.5 Use informal arguments about angle sum, exterior angles, and angles formed when parallel lines are cut by a transversal (Proving BA is parallel to CE from supplementary same-side interior angles, transferring parallelogram angles, and finding the isosceles base angles)

⭐ When a shape has several equal sides, build an equilateral triangle on one of them: it copies that length and drops a clean 60 degree angle that can turn a messy corner into parallel lines you can actually use.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Build an equilateral triangle on CD

Slice the 170 degree corner

Spot a pair of parallel sides

Cash in the parallelogram

Read the isosceles triangle AED

Subtract to reach angle BAD

Review

CCSS standards used (min grade 8)

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