AMC 8 · 2000 · #24
Grade 7 geometry-2dProblem
If ∠A=20∘ and ∠AFG=∠AGF, then ∠B+∠D=
Pick an answer.
AMC 8 2000 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 the figure, points $A$, $G$, $F$ form a triangle with $\angle A = 20^\circ$ and equal base angles $\angle AFG = \angle AGF$. Lines through $F$ also create triangle $BFD$. Find $\angle B + \angle D$.
Givens: $\angle A = 20^\circ$ in triangle $AGF$; $\angle AFG = \angle AGF$ (triangle $AGF$ is isosceles at $A$); $F$ lies on segment $AD$ (so $A$, $F$, $D$ are collinear); $F$ lies on segment $BE$, with $G$ on the same ray from $F$ as $B$ (so ray $FG$ = ray $FB$); Answer choices: (A) $48^\circ$, (B) $60^\circ$, (C) $72^\circ$, (D) $80^\circ$, (E) $90^\circ$
Unknowns: The sum $\angle B + \angle D$
Understand
Restated: In the figure, points $A$, $G$, $F$ form a triangle with $\angle A = 20^\circ$ and equal base angles $\angle AFG = \angle AGF$. Lines through $F$ also create triangle $BFD$. Find $\angle B + \angle D$.
Givens: $\angle A = 20^\circ$ in triangle $AGF$; $\angle AFG = \angle AGF$ (triangle $AGF$ is isosceles at $A$); $F$ lies on segment $AD$ (so $A$, $F$, $D$ are collinear); $F$ lies on segment $BE$, with $G$ on the same ray from $F$ as $B$ (so ray $FG$ = ray $FB$); Answer choices: (A) $48^\circ$, (B) $60^\circ$, (C) $72^\circ$, (D) $80^\circ$, (E) $90^\circ$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram
The figure looks busy, but the relevant information lives in just two triangles: $AGF$ (where $\angle A = 20^\circ$ and the base angles are equal) and $BFD$ (whose interior angles include $\angle B$ and $\angle D$). Tool #7 (Identify Subproblems) splits the figure along these two triangles, linked at point $F$. Tool #1 (Draw a Diagram) is the sidekick: mark the known $20^\circ$ at $A$, label the matching base angles, then use the straight line $AFD$ to carry that information from one triangle to the other. No algebra, just the triangle-angle-sum rule applied twice.
Execute — Answer: D
7.G.B.5 Step 1 - Solve subproblem 1: find the base angle in triangle $AGF$.
- The three angles must add to $180^\circ$, and the two base angles are equal, so call each one $x$.
💡 An isosceles triangle with one $20^\circ$ vertex angle must have two $80^\circ$ base angles.
7.G.B.5 Step 2 - Transfer that result across point $F$.
- Because $A$, $F$, $D$ lie on one straight line, $\angle AFG$ and $\angle GFD$ form a linear pair and add to $180^\circ$.
💡 The straight segment $AD$ acts as a bridge — whatever angle sits on one side of $F$ leaves $180^\circ$ minus that angle on the other side.
4.G.A.1 Step 3 - Identify the angle of triangle $BFD$ at vertex $F$.
- Point $G$ lies on ray $FB$ (both are on the same side of $F$ along line $BE$), so the ray $FG$ and ray $FB$ point in the same direction.
- That means $\angle BFD$ and $\angle GFD$ are the very same angle.
💡 Replacing $G$ with $B$ on the same ray doesn't change the angle — angles depend on the ray's direction, not on which point you name on it.
7.G.B.5 Step 4 Solve subproblem 2: apply the triangle-angle-sum rule to triangle $BFD$.
💡 Once the third angle of a triangle is pinned down, the sum of the other two is forced — even without knowing each one individually.
7.G.B.5 Solve subproblem 1: find the base angle in triangle $AGF$. The three angles must 7.G.B.5 Transfer that result across point $F$. Because $A$, $F$, $D$ lie on one straight 4.G.A.1 Identify the angle of triangle $BFD$ at vertex $F$. Point $G$ lies on ray $FB$ ( 7.G.B.5 Solve subproblem 2: apply the triangle-angle-sum rule to triangle $BFD$. Review
Reasonableness: The two unknown angles $\angle B$ and $\angle D$ can each take many values (the figure doesn't fix them separately), yet the problem expects a single number — that's already a hint that only the *sum* is determined, exactly what the triangle-angle-sum rule provides once $\angle BFD$ is known. The arithmetic checks: $20 + 80 + 80 = 180$ in triangle $AGF$, $80 + 100 = 180$ on line $AD$, and $\angle B + \angle D + 100 = 180$ in triangle $BFD$, giving $80^\circ$ — choice (D). The answer is also positive and less than $180^\circ$, as any pair of triangle angles must be.
Alternative: Tool #5 (Look for a Pattern) via the exterior-angle shortcut: $\angle AFG$ is the exterior angle of triangle $BFD$ at $F$ (because $A$ is on the extension of segment $DF$ past $F$). The exterior-angle theorem says an exterior angle equals the sum of the two remote interior angles, so $\angle B + \angle D = \angle AFG = 80^\circ$ in a single step — same answer (D), one line of work.
CCSS standards used (min grade 7)
7.G.B.5Use facts about supplementary, complementary, vertical, and adjacent angles in a multi-step problem to write and solve simple equations for an unknown angle in a figure (Applying the triangle-angle-sum rule in triangles $AGF$ and $BFD$, and using the supplementary pair $\angle AFG + \angle GFD = 180^\circ$ to carry information across the shared vertex $F$.)4.G.A.1Draw points, lines, line segments, rays, angles, and identify these in two-dimensional figures (Recognizing that ray $FG$ and ray $FB$ point in the same direction along line $BE$, which lets us replace $\angle GFD$ with the triangle-$BFD$ angle $\angle BFD$.)
⭐ The figure looks tangled, but only two triangles do the work: $AGF$ on top and $BFD$ on the bottom, meeting at $F$. The isosceles top triangle pins $\angle AFG = 80^\circ$; the straight line through $F$ flips that to $\angle BFD = 100^\circ$; and the triangle-angle-sum rule in $BFD$ forces $\angle B + \angle D = 80^\circ$ — choice (D).
⭐ The figure looks tangled, but only two triangles do the work: $AGF$ on top and $BFD$ on the bottom, meeting at $F$. The isosceles top triangle pins $\angle AFG = 80^\circ$; the straight line through $F$ flips that to $\angle BFD = 100^\circ$; and the triangle-angle-sum rule in $BFD$ forces $\angle B + \angle D = 80^\circ$ — choice (D).
More like this
Same archetype — closest grade level first.