AMC 10 · 2022 · #20
Grade 8 geometry-2dProblem
Let ABCD be a rhombus with ∠ADC=46∘. Let E be the midpoint of CD, and let F be the point
on BE such that AF is perpendicular to BE. What is the degree measure of ∠BFC?
Pick an answer.
AMC 10 2022 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 rhombus $ABCD$ with $\angle ADC = 46°$, let $E$ be the midpoint of $\overline{CD}$. Let $F$ on $\overline{BE}$ satisfy $\overline{AF} \perp \overline{BE}$. Find $\angle BFC$.
Givens: $ABCD$ is a rhombus (all four sides equal); $\angle ADC = 46°$; $E$ is the midpoint of $\overline{CD}$; $F$ on $\overline{BE}$ with $\overline{AF} \perp \overline{BE}$; Choices: (A) $110$, (B) $111$, (C) $112$, (D) $113$, (E) $114$
Unknowns: The degree measure of $\angle BFC$
Understand
Restated: In rhombus $ABCD$ with $\angle ADC = 46°$, let $E$ be the midpoint of $\overline{CD}$. Let $F$ on $\overline{BE}$ satisfy $\overline{AF} \perp \overline{BE}$. Find $\angle BFC$.
Givens: $ABCD$ is a rhombus (all four sides equal); $\angle ADC = 46°$; $E$ is the midpoint of $\overline{CD}$; $F$ on $\overline{BE}$ with $\overline{AF} \perp \overline{BE}$; Choices: (A) $110$, (B) $111$, (C) $112$, (D) $113$, (E) $114$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #11 Work Backwards
Tool #1 (Diagram): Sketch the rhombus, mark $E$ on $\overline{CD}$, and extend line $\overline{AD}$ until it meets line $\overline{BE}$ at a point $G$. The extension is the unlock — once $G$ exists, two parallel-line triangles ($\triangle GDE \sim \triangle GAB$ at ratio $1:2$) hand us $D$ as the midpoint of $\overline{AG}$. Tool #7 (Subproblems): break the answer into (i) prove $FD = AD$ via the median-to-hypotenuse theorem in right $\triangle AFG$, (ii) show $F, A, G, C$ lie on a circle centered at $D$, (iii) use the inscribed-angle theorem to convert $\angle GFC$ from a central angle. Tool #11 (Work Backwards): the target $\angle BFC$ is supplementary to $\angle GFC$ (since $B, F, G$ are collinear), so we work backwards from $\angle ADC = 46°$ to $\angle GDC = 134°$ to $\angle GFC = 67°$ to $\angle BFC = 113°$.
Execute — Answer: D
8.G.A.5 Step 1 - Extend ray $\overline{AD}$ beyond $D$ and ray $\overline{BE}$ beyond $E$ until they meet at a point $G$.
- Since $ABCD$ is a rhombus, $\overline{AB} \parallel \overline{CD}$, so $\overline{AB} \parallel \overline{DE}$ (because $E$ lies on $\overline{CD}$).
- Triangles $\triangle GDE$ and $\triangle GAB$ share vertex $G$ and have $\overline{DE} \parallel \overline{AB}$, so they're similar (AA).
💡 Extend two lines until they meet; parallel sides force the two triangles to be similar.
7.G.A.1 Step 2 - The similarity ratio is $\frac{DE}{AB} = \frac{s/2}{s} = \frac{1}{2}$ where $s$ is the rhombus side ($E$ being the midpoint of $\overline{CD}$).
- So $\frac{GD}{GA} = \frac{1}{2}$, which means $GA = 2 \cdot GD$ — equivalently, $D$ is the midpoint of segment $\overline{AG}$.
💡 Half-as-long $\overline{DE}$ vs $\overline{AB}$ means $D$ sits exactly halfway between $A$ and $G$ on the slanted line.
7.G.B.5 Step 3 - Since $F$ is on line $\overline{BE}$ and $\overline{AF} \perp \overline{BE}$, in fact $\overline{AF} \perp \overline{AG}$'s extension...
- wait — $F$ lies on the line through $B$ and $E$, which is the same as the line through $B, E, G$.
- So $\overline{AF}$ is perpendicular to that line at $F$, making $\angle AFG = 90°$.
- So $\triangle AFG$ is a right triangle with hypotenuse $\overline{AG}$ and right angle at $F$.
💡 $F$ lives on the extended line $\overline{BG}$, so the perpendicular from $A$ lands at a right angle to the hypotenuse $\overline{AG}$.
7.G.A.2 Step 4 - Apply the median-to-hypotenuse theorem: in a right triangle, the median from the right-angle vertex to the hypotenuse has length equal to half the hypotenuse.
- Here $D$ is the midpoint of hypotenuse $\overline{AG}$, so $\overline{FD}$ is that median.
- Therefore $FD = \tfrac{1}{2} AG = AD = DG$.
💡 Median from the right-angle vertex to the hypotenuse is exactly half the hypotenuse — so $F$, $A$, $G$ are equidistant from $D$.
7.G.B.4 Step 5 - Also $AD = CD$ (rhombus sides equal), so $CD = AD = DG = FD$.
- Therefore the four points $F, A, G, C$ all lie at distance $AD$ from $D$ — they lie on a circle centered at $D$ with radius $AD$.
💡 Four points at the same distance from $D$ — that's exactly the definition of a circle through them.
7.G.B.5 Step 6 - Use the inscribed-angle theorem.
- The inscribed angle $\angle GFC$ subtends the same arc $\widehat{GC}$ as the central angle $\angle GDC$, so $\angle GFC = \tfrac{1}{2} \angle GDC$.
- Now $A$, $D$, $G$ are collinear (we built $G$ as the extension of $\overline{AD}$), so $\angle GDC$ and $\angle ADC$ are supplementary: $\angle GDC = 180° - 46° = 134°$.
- Hence $\angle GFC = \tfrac{1}{2}(134°) = 67°$.
💡 Inscribed angle is half the central angle on the same arc; supplementary angles fix the central one.
7.G.B.5 Step 7 - Finally use Tool #11 (Work Backwards) on the target: $B$, $F$, $G$ are collinear (they all lie on line $\overline{BE}$ extended), so $\angle BFC$ and $\angle GFC$ are supplementary.
- $\angle BFC = 180° - 67° = 113°$.
- Answer $\textbf{(D)}$.
💡 Straight line through $B, F, G$ — the two angles on either side at $F$ sum to $180°$.
8.G.A.5 Extend ray $\overline{AD}$ beyond $D$ and ray $\overline{BE}$ beyond $E$ until t 7.G.A.1 The similarity ratio is $\frac{DE}{AB} = \frac{s/2}{s} = \frac{1}{2}$ where $s$ 7.G.B.5 Since $F$ is on line $\overline{BE}$ and $\overline{AF} \perp \overline{BE}$, in 7.G.A.2 Apply the median-to-hypotenuse theorem: in a right triangle, the median from the 7.G.B.4 Also $AD = CD$ (rhombus sides equal), so $CD = AD = DG = FD$. Therefore the four 7.G.B.5 Use the inscribed-angle theorem. The inscribed angle $\angle GFC$ subtends the s 7.G.B.5 Finally use Tool #11 (Work Backwards) on the target: $B$, $F$, $G$ are collinear Review
Reasonableness: Sanity: $\angle ADC = 46°$, and the answer $113° = 90° + 23° = 90° + \tfrac{46°}{2}$. That clean relation $\angle BFC = 90° + \tfrac{1}{2} \angle ADC$ matches the inscribed-angle / supplementary-angle chain: $180° - \tfrac{1}{2}(180° - 46°) = 180° - 90° + 23° = 113°$. The choices are $110, 111, 112, 113, 114$ — only $113$ fits the half-angle formula, and the others tempt off-by-one slips in the supplementary step.
Alternative: Tool #13 (Convert to Algebra) with coordinates: place $D$ at origin, $C$ on a ray making angle $46°$ with $\overline{DA}$, set $s = 2$ so $E$ is at half-position on $\overline{DC}$. Compute $\overline{BE}$ as a line, drop a perpendicular from $A$ to find $F$, then evaluate $\angle BFC$ using the dot product. Slogs through trig but lands on $113°$ — the synthetic median-to-hypotenuse path is much cleaner.
CCSS standards used (min grade 8)
8.G.A.5Use informal arguments to establish facts about angle sum and exterior angles (Setting up the AA similarity between $\triangle GDE$ and $\triangle GAB$ via parallel sides and shared vertex.)7.G.A.1Solve problems involving scale drawings of geometric figures (Using the $1:2$ similarity ratio to conclude $D$ is the midpoint of $\overline{AG}$.)7.G.B.4Know the formulas for area and circumference of a circle (Identifying the four points $F, A, G, C$ as equidistant from $D$, hence lying on a circle.)7.G.B.5Use facts about supplementary, complementary, vertical, and adjacent angles (Supplementary pairs $\angle ADC + \angle GDC = 180°$ and $\angle GFC + \angle BFC = 180°$, plus the inscribed-angle / right-triangle chain.)7.G.A.2Draw geometric shapes with given conditions including triangles (Applying the median-to-hypotenuse theorem inside right $\triangle AFG$ to get $FD = AD$.)
⭐ Extend $\overline{AD}$ and $\overline{BE}$ to meet at $G$. Then $D$ is the midpoint of the hypotenuse of right $\triangle AFG$, so $FD = AD = CD$ — four points $F, A, G, C$ on a circle centered at $D$. The inscribed angle is half the central one, $\angle GFC = \tfrac{1}{2}(134°) = 67°$, and the supplement gives $\angle BFC = 113°$, choice $\textbf{(D)}$.
⭐ Extend $\overline{AD}$ and $\overline{BE}$ to meet at $G$. Then $D$ is the midpoint of the hypotenuse of right $\triangle AFG$, so $FD = AD = CD$ — four points $F, A, G, C$ on a circle centered at $D$. The inscribed angle is half the central one, $\angle GFC = \tfrac{1}{2}(134°) = 67°$, and the supplement gives $\angle BFC = 113°$, choice $\textbf{(D)}$.