AMC 8 · 2014 · #9
Grade 8 geometry-2dProblem
In △ABC, D is a point on side AC such that BD=DC and ∠BCD measures 70∘. What is the degree measure of ∠ADB?
Pick an answer.
AMC 8 2014 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.
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 $\angle ADB$.
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)
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
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 $\angle DBC = \angle BCD = 70^\circ$.
💡 Classifying $\triangle BDC$ as isosceles from its equal side marks is the Grade 5 "classify 2D figures by properties" move.
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 $\angle BDC$.
💡 "The three interior angles of a triangle add to $180^\circ$" is the Grade 8 informal angle-sum fact.
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 $180^\circ$.
💡 Recognizing a linear pair (supplementary adjacent angles on a straight line) is the Grade 7 angle-relationship skill.
5.G.B.4 In the diagram, $\triangle BDC$ has $BD = DC$, so it is isosceles with base $\ov 8.G.A.5 The interior angles of any triangle add to $180^\circ$. In $\triangle BDC$, two 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.4Classify 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.5Use 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.5Use 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.7Recognize 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 $180^\circ$.
⭐ 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 $180^\circ$.
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