AMC 10 · 2024 · #10
Grade 8 geometry-2dProblem
Quadrilateral ABCD is a parallelogram, and E is the midpoint of the side AD. Let F be the intersection of lines EB and AC. What is the ratio of the area of
quadrilateral CDEF to the area of △CFB?
Pick an answer.
AMC 10 2024 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: $ABCD$ is a parallelogram. $E$ is the midpoint of side $\overline{AD}$, and $F$ is the point where line $EB$ crosses diagonal $AC$. Find the ratio $\text{area}(CDEF) : \text{area}(\triangle CFB)$.
Givens: $ABCD$ is a parallelogram, so $\overline{AD} \parallel \overline{BC}$ and $AD = BC$; $E$ is the midpoint of $\overline{AD}$, so $AE = \tfrac{1}{2} AD = \tfrac{1}{2} BC$; $F = \overline{EB} \cap \overline{AC}$; Answer choices: (A) $5:4$, (B) $4:3$, (C) $3:2$, (D) $5:3$, (E) $2:1$
Unknowns: The ratio of $\text{area}(CDEF)$ to $\text{area}(\triangle CFB)$
Understand
Restated: $ABCD$ is a parallelogram. $E$ is the midpoint of side $\overline{AD}$, and $F$ is the point where line $EB$ crosses diagonal $AC$. Find the ratio $\text{area}(CDEF) : \text{area}(\triangle CFB)$.
Givens: $ABCD$ is a parallelogram, so $\overline{AD} \parallel \overline{BC}$ and $AD = BC$; $E$ is the midpoint of $\overline{AD}$, so $AE = \tfrac{1}{2} AD = \tfrac{1}{2} BC$; $F = \overline{EB} \cap \overline{AC}$; Answer choices: (A) $5:4$, (B) $4:3$, (C) $3:2$, (D) $5:3$, (E) $2:1$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
The setup is purely geometric — a parallelogram, a midpoint, a diagonal, and one intersection point. Tool #1 (Draw a Diagram) puts every length and triangle on paper so similar-triangle pairs can be spotted at a glance. Tool #7 (Identify Subproblems) then breaks the answer into bite-sized pieces: (a) find the similarity ratio of $\triangle AEF$ and $\triangle CBF$, (b) use that ratio to express the areas of $\triangle AEF, \triangle CBF, \triangle ABF$ in terms of one variable $x$, (c) use the diagonal to get the area of $\triangle ADC$, (d) subtract to get $\text{area}(CDEF)$, then form the requested ratio.
Execute — Answer: A
8.G.A.5 Step 1 - Identify the similar triangles in the diagram.
- Lines $\overline{AD}$ and $\overline{BC}$ are parallel, so the transversal $\overline{AC}$ gives $\angle FAE = \angle FCB$ and the transversal $\overline{EB}$ gives $\angle FEA = \angle FBC$.
- Combined with the vertical angle at $F$, the triangles share all three angles.
💡 Parallel lines cut by transversals force matching alternate-interior angles — Grade 8 informal angle arguments make the AA similarity instant.
8.G.A.4 Step 2 - Subproblem A: find the similarity ratio.
- Since $E$ is the midpoint of $\overline{AD}$, $AE = \tfrac{1}{2} AD = \tfrac{1}{2} BC$.
- Corresponding sides $AE$ and $CB$ are in ratio $1:2$, so all corresponding lengths share this ratio.
💡 Similar figures scale every length by the same factor — Grade 8 similarity. $E$ being a midpoint pins the factor at $\tfrac{1}{2}$.
8.G.A.4 Step 3 - Subproblem B: turn the length ratio into area ratios.
- Areas of similar triangles scale as the square of the length ratio.
- Let $[\triangle AEF] = x$.
- Then $[\triangle CBF] = 4x$.
- Also, $\triangle ABF$ and $\triangle CBF$ share the apex $B$ above bases $\overline{AF}$ and $\overline{FC}$, so their areas scale with the base ratio $AF:FC = 1:2$, giving $[\triangle ABF] = 2x$.
💡 Ratio-of-areas = (ratio-of-sides)$^2$ for similar triangles, and same-height triangles split area by base — two Grade 8 similarity ideas in one step.
6.G.A.1 Step 4 - Subproblem C: find $[\triangle ADC]$ via the diagonal.
- Diagonal $\overline{AC}$ splits the parallelogram into two congruent triangles, so $[\triangle ABC] = [\triangle ADC]$.
- From $[\triangle ABC] = [\triangle ABF] + [\triangle CBF] = 2x + 4x = 6x$, we get $[\triangle ADC] = 6x$.
💡 A diagonal of a parallelogram halves its area into two congruent triangles — a Grade 6 "area by composing" fact.
6.G.A.1 Step 5 - Subproblem D: compute the requested ratio.
- Quadrilateral $CDEF$ is $\triangle ADC$ with $\triangle AEF$ cut off, so $[CDEF] = 6x - x = 5x$.
- Compare to $[\triangle CBF] = 4x$.
💡 Areas can be added and subtracted because the region $CDEF$ is just $\triangle ADC$ minus $\triangle AEF$ — Grade 6 "compose and decompose polygons."
8.G.A.5 Identify the similar triangles in the diagram. Lines $\overline{AD}$ and $\overl 8.G.A.4 Subproblem A: find the similarity ratio. Since $E$ is the midpoint of $\overline 8.G.A.4 Subproblem B: turn the length ratio into area ratios. Areas of similar triangles 6.G.A.1 Subproblem C: find $[\triangle ADC]$ via the diagonal. Diagonal $\overline{AC}$ 6.G.A.1 Subproblem D: compute the requested ratio. Quadrilateral $CDEF$ is $\triangle AD Review
Reasonableness: Sanity check the parts: the parallelogram has total area $12x$ ($6x + 6x$), and the four named pieces inside it — $\triangle AEF$ ($x$), $\triangle CBF$ ($4x$), $\triangle ABF$ ($2x$), and quadrilateral $CDEF$ ($5x$) — sum to $x + 4x + 2x + 5x = 12x$, accounting for the whole parallelogram exactly. The ratio $5:4$ is also the only choice strictly between $1:1$ and $3:2$, matching the geometric intuition that $CDEF$ is just a bit bigger than $\triangle CBF$.
Alternative: Tool #6 (Guess and Check) using coordinates: place $A = (0,0)$, $B = (2,0)$, $C = (3,2)$, $D = (1,2)$. Then $E$, midpoint of $\overline{AD}$, is $(1/2, 1)$. Diagonal $\overline{AC}$ has the equation $y = \tfrac{2}{3} x$. Line $\overline{EB}$ passes through $(1/2, 1)$ and $(2, 0)$. Solving the two gives $F = (1, 2/3)$. The shoelace formula yields $[CDEF] = 5/3$ and $[\triangle CBF] = 4/3$, and $\tfrac{5/3}{4/3} = \tfrac{5}{4}$ — the same answer (A).
CCSS standards used (min grade 8)
8.G.A.5Use informal arguments to establish facts about angle sum and exterior angles (Using alternate-interior angles from $\overline{AD} \parallel \overline{BC}$ (with transversals $\overline{AC}$ and $\overline{EB}$) and vertical angles at $F$ to show $\triangle AEF \sim \triangle CBF$ by the angle-angle criterion.)8.G.A.4Understand that a two-dimensional figure is similar to another using transformations (Using $AE = \tfrac{1}{2} BC$ to set the similarity ratio at $\tfrac{1}{2}$ and then concluding $[\triangle AEF] : [\triangle CBF] = 1:4$ from the square of the length ratio.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Adding $[\triangle ABF] + [\triangle CBF] = 6x$ to get $[\triangle ABC]$, recognizing $[\triangle ABC] = [\triangle ADC]$ via the diagonal, and subtracting $[\triangle AEF]$ from $[\triangle ADC]$ to get $[CDEF] = 5x$.)
⭐ This AMC 10 problem only needs Grade 8 similar-triangle reasoning — the midpoint forces a $1:2$ side ratio, which becomes a $1:4$ area ratio, and the rest is adding and subtracting triangle areas!
⭐ This AMC 10 problem only needs Grade 8 similar-triangle reasoning — the midpoint forces a $1:2$ side ratio, which becomes a $1:4$ area ratio, and the rest is adding and subtracting triangle areas!
More like this
Same archetype — closest grade level first.
