AMC 8 · 2014 · #9

Easy mode Grade 8

Problem

Picture a triangle with corners $A$ , $B$ , and $C$ .

Pick a point $D$ somewhere on side $\overline{AC}$ . Then draw a line from $B$ to $D$ . Make sure that $BD$ and $DC$ have the same length, so triangle $BDC$ has two equal sides.

The corner of the small triangle at $C$ measures $70^\circ$ .

What is the size of $\angle ADB$ (the angle at $D$ on the other side)?

$\textbf{(A) }100\qquad\textbf{(B) }120\qquad\textbf{(C) }135\qquad\textbf{(D) }140\qquad \textbf{(E) }150$

Pick an answer.

(A)

100

(B)

120

(C)

135

(D)

140

(E)

150

View mode:

Toolkit + CCSS Solution

Understand

Restated: In $\triangle ABC$, point $D$ sits on side $\overline{AC}$ so that $BD = DC$, and $\angle BCD = 70^\circ$. Find the degree measure of $\angle ADB$.

Givens: $D$ is a point on side $\overline{AC}$ of $\triangle ABC$; $BD = DC$ (two sides of $\triangle BDC$ are equal); $\angle BCD = 70^\circ$; Answer choices: (A) $100$, (B) $120$, (C) $135$, (D) $140$, (E) $150$ (degrees)

Unknowns: The degree measure of $\angle ADB$

Understand

Restated: In $\triangle ABC$, point $D$ sits on side $\overline{AC}$ so that $BD = DC$, and $\angle BCD = 70^\circ$. Find the degree measure of $\angle ADB$.

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems

The problem is purely geometric, so Tool #1 (Draw a Diagram) is the starting move — sketch $\triangle ABC$, mark $D$ on $\overline{AC}$, draw $\overline{BD}$, and tick the two equal sides $BD$ and $DC$ to make the isosceles structure of $\triangle BDC$ visible. Once the picture is in place, Tool #7 (Identify Subproblems) splits the angle hunt into two clean steps: (a) find $\angle BDC$ inside the isosceles triangle $\triangle BDC$, then (b) use the fact that $\angle ADB$ and $\angle BDC$ form a straight line at $D$ to get $\angle ADB$. Each sub-step is a one-line angle calculation.

Execute — Answer: D

#1 Draw a Diagram 5.G.B.4 Step 1

In the diagram, $\triangle BDC$ has $BD = DC$, so it is isosceles with base $\overline{BC}$.
The two base angles — the ones across from the equal sides — must be equal.
The angle across from $BD$ is $\angle BCD$, and the angle across from $DC$ is $\angle DBC$, so $\angle DBC = \angle BCD = 70^\circ$.

$$\angle DBC = \angle BCD = 70^\circ$$

💡 Classifying $\triangle BDC$ as isosceles from its equal side marks is the Grade 5 "classify 2D figures by properties" move.

#7 Identify Subproblems 8.G.A.5 Step 2

The interior angles of any triangle add to $180^\circ$.
In $\triangle BDC$, two of the three angles are already known ($70^\circ$ each), so subtract their sum from $180^\circ$ to get the third angle $\angle BDC$.

$$\angle BDC = 180^\circ - 70^\circ - 70^\circ = 40^\circ$$

💡 "The three interior angles of a triangle add to $180^\circ$" is the Grade 8 informal angle-sum fact.

#7 Identify Subproblems 7.G.B.5 Step 3

Because $D$ lies on segment $\overline{AC}$, the rays $\overrightarrow{DA}$ and $\overrightarrow{DC}$ point in opposite directions and together make a straight line.
The angles $\angle ADB$ and $\angle BDC$ are the two pieces of that straight line on either side of $\overline{DB}$, so they are supplementary — their measures add to $180^\circ$.

$$\angle ADB + \angle BDC = 180^\circ \;\Rightarrow\; \angle ADB = 180^\circ - 40^\circ = 140^\circ \;\Rightarrow\; \textbf{(D)}$$

💡 Recognizing a linear pair (supplementary adjacent angles on a straight line) is the Grade 7 angle-relationship skill.

[1] #1 5.G.B.4 In the diagram, $\triangle BDC$ has $BD = DC$, so it is isosceles with base $\ov

[2] #7 8.G.A.5 The interior angles of any triangle add to $180^\circ$. In $\triangle BDC$, two

[3] #7 7.G.B.5 Because $D$ lies on segment $\overline{AC}$, the rays $\overrightarrow{DA}$ and

Review

Reasonableness: Check the additivity directly: $\angle ADB + \angle BDC = 140^\circ + 40^\circ = 180^\circ$, which matches the straight line $\overline{AC}$ at $D$. Also, $140^\circ$ is obtuse, which fits the picture — $\overline{DB}$ leans away from $\overline{DC}$ (toward $A$), so the angle on the $A$-side at $D$ should clearly be the larger of the two pieces. The two base angles of $\triangle BDC$ are $70^\circ$ each, and $70^\circ + 70^\circ + 40^\circ = 180^\circ$ also checks out — every angle in sight adds up correctly.

Alternative: Tool #6 (Guess and Check) using the exterior-angle shortcut. The angle $\angle ADB$ is the exterior angle of $\triangle BDC$ at vertex $D$, and the exterior-angle theorem says it equals the sum of the two non-adjacent interior angles of that triangle: $\angle ADB = \angle DBC + \angle BCD = 70^\circ + 70^\circ = 140^\circ$. Same answer in one line — and it rules out every other choice immediately.

CCSS standards used (min grade 8)

5.G.B.4 Classify two-dimensional figures in a hierarchy based on properties (Recognizing $\triangle BDC$ as isosceles from the equal-sides condition $BD = DC$, which forces the base angles $\angle DBC$ and $\angle BCD$ to be equal.)
8.G.A.5 Use informal arguments to establish facts about the angle sum of triangles (Applying the triangle interior angle-sum fact ($180^\circ$) inside $\triangle BDC$ to compute $\angle BDC = 180^\circ - 70^\circ - 70^\circ = 40^\circ$.)
7.G.B.5 Use facts about supplementary, complementary, vertical, and adjacent angles to find unknown angles (Recognizing $\angle ADB$ and $\angle BDC$ as a linear pair on segment $\overline{AC}$, so $\angle ADB + \angle BDC = 180^\circ$.)
4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems (Doing the angle arithmetic — adding $70^\circ + 70^\circ$ and subtracting from $180^\circ$ — to find both $\angle BDC$ and $\angle ADB$.)

⭐ This AMC 8 problem only needs the Grade 8 fact that a triangle's three angles add to $180^\circ$ — plus the Grade 7 fact that two angles on a straight line add to $180^\circ$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 8)

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