AMC 10 · 2019 · #13
Grade 8 geometry-2dProblem
Let △ABC be an isosceles triangle with BC=AC and ∠ACB=40∘. Construct the circle with diameter BC, and let D and E be the other intersection points of the circle with the sides AC and AB, respectively. Let F be the intersection of the diagonals of the quadrilateral BCDE. What is the degree measure of \angle BFC ?
Pick an answer.
AMC 10 2019 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: $\triangle ABC$ is isosceles with $BC = AC$ and $\angle ACB = 40^{\circ}$. A circle is drawn with diameter $\overline{BC}$. It meets $\overline{AC}$ again at $D$ and $\overline{AB}$ again at $E$. The diagonals of quadrilateral $BCDE$ are $\overline{BD}$ and $\overline{CE}$; they meet at $F$. Find $\angle BFC$.
Givens: $\triangle ABC$ isosceles with $BC = AC$; $\angle ACB = 40^{\circ}$; Circle has diameter $\overline{BC}$; $D$ on $\overline{AC}$, $E$ on $\overline{AB}$, both on the circle; $F$ is the intersection of $\overline{BD}$ and $\overline{CE}$ inside the triangle; Answer choices: $90^{\circ}, 100^{\circ}, 105^{\circ}, 110^{\circ}, 120^{\circ}$
Unknowns: $\angle BFC$ in degrees
Understand
Restated: $\triangle ABC$ is isosceles with $BC = AC$ and $\angle ACB = 40^{\circ}$. A circle is drawn with diameter $\overline{BC}$. It meets $\overline{AC}$ again at $D$ and $\overline{AB}$ again at $E$. The diagonals of quadrilateral $BCDE$ are $\overline{BD}$ and $\overline{CE}$; they meet at $F$. Find $\angle BFC$.
Givens: $\triangle ABC$ isosceles with $BC = AC$; $\angle ACB = 40^{\circ}$; Circle has diameter $\overline{BC}$; $D$ on $\overline{AC}$, $E$ on $\overline{AB}$, both on the circle; $F$ is the intersection of $\overline{BD}$ and $\overline{CE}$ inside the triangle; Answer choices: $90^{\circ}, 100^{\circ}, 105^{\circ}, 110^{\circ}, 120^{\circ}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
Tool #1 (Draw): sketch $\triangle ABC$ with the circle on diameter $\overline{BC}$, and mark the inscribed right angles at $D$ and $E$ (Thales). Tool #7 (Subproblems): peel off three tiny angle-chasing triangles in order — $\triangle ABC$ gives the base angles; $\triangle BEC$ (right-angled at $E$) gives $\angle BCE$; $\triangle BDC$ (right-angled at $D$) gives $\angle DBC$; finally $\triangle BFC$ adds up to $180^{\circ}$. Tool #3 matches $110^{\circ}$ to the choices.
Execute — Answer: D
4.G.A.1 Step 1 - Sketch the figure.
- Draw $\triangle ABC$ with $A$ at the top and $BC$ at the bottom.
- Draw the circle whose diameter is $\overline{BC}$ — its center is the midpoint of $\overline{BC}$.
- Mark $D$ where the circle meets $\overline{AC}$ a second time, and $E$ where the circle meets $\overline{AB}$.
- The diagonals of $BCDE$ are $\overline{BD}$ and $\overline{CE}$; mark their crossing point $F$.
💡 Grade 4 figures: a clean labeled picture makes every angle visible.
8.G.A.5 Step 2 - Find the base angles of $\triangle ABC$.
- The triangle is isosceles with $BC = AC$, so the angles opposite the equal sides are equal: $\angle BAC = \angle ABC$.
- The triangle's angles sum to $180^{\circ}$, so $\angle BAC + \angle ABC = 180^{\circ} - 40^{\circ} = 140^{\circ}$, giving $\angle BAC = \angle ABC = 70^{\circ}$.
💡 Grade 8 angle sums: the two equal base angles split the leftover $140^{\circ}$ evenly.
7.G.B.5 Step 3 - Use Thales' theorem at $E$.
- Because $\overline{BC}$ is the diameter and $E$ is on the circle, the inscribed angle $\angle BEC$ subtends the diameter and is $90^{\circ}$.
- Look at $\triangle BEC$: its angles are $\angle BEC = 90^{\circ}$ and $\angle EBC = \angle ABC = 70^{\circ}$, so the third angle $\angle BCE = 180^{\circ} - 90^{\circ} - 70^{\circ} = 20^{\circ}$.
💡 Grade 7 angle facts: a right angle and a known angle force the third in a triangle.
7.G.B.5 Step 4 - Use Thales at $D$.
- Because $D$ is on the circle and $\overline{BC}$ is the diameter, $\angle BDC = 90^{\circ}$.
- In $\triangle BDC$: $\angle BDC = 90^{\circ}$ and $\angle DCB = \angle ACB = 40^{\circ}$, so $\angle DBC = 180^{\circ} - 90^{\circ} - 40^{\circ} = 50^{\circ}$.
💡 Grade 7 angle facts: another right angle peels off another tiny triangle.
8.G.A.5 Step 5 - Now look at $\triangle BFC$.
- The angle at $B$ is $\angle FBC$, which is the same ray-pair as $\angle DBC$ because $F$ lies on $\overline{BD}$; so $\angle FBC = 50^{\circ}$.
- The angle at $C$ is $\angle FCB$, which is $\angle ECB = 20^{\circ}$ because $F$ lies on $\overline{CE}$.
- Sum to $180^{\circ}$: $\angle BFC = 180^{\circ} - 50^{\circ} - 20^{\circ} = 110^{\circ}$.
💡 Grade 8: the diagonal-intersection triangle is just one more angle-sum.
4.MD.C.6 Step 6 Match $110^{\circ}$ to the choices: $(D)$.
💡 Grade 4 angle measure: pick the matching degree value.
4.G.A.1 Sketch the figure. Draw $\triangle ABC$ with $A$ at the top and $BC$ at the bott 8.G.A.5 Find the base angles of $\triangle ABC$. The triangle is isosceles with $BC = AC 7.G.B.5 Use Thales' theorem at $E$. Because $\overline{BC}$ is the diameter and $E$ is o 7.G.B.5 Use Thales at $D$. Because $D$ is on the circle and $\overline{BC}$ is the diame 8.G.A.5 Now look at $\triangle BFC$. The angle at $B$ is $\angle FBC$, which is the same 4.MD.C.6 Match $110^{\circ}$ to the choices: $(D)$. Review
Reasonableness: Since $\angle BFC$ and $\angle DFE$ are vertical angles at $F$, and the other pair of vertical angles is $\angle BFE = \angle DFC = 70^{\circ}$ (so they sum to $360^{\circ}$ around $F$), the four angles at $F$ are $110^{\circ}, 70^{\circ}, 110^{\circ}, 70^{\circ}$. That fits a typical convex quadrilateral with its diagonals crossing inside. $110^{\circ}$ is also obtuse, which matches the picture — $F$ sits a bit below the top vertex $A$, so $\angle BFC$ should open wider than $90^{\circ}$.
Alternative: Tool #4 (use a known cyclic-quadrilateral identity): $BCDE$ is cyclic because all four points lie on the circle, so the exterior angle at $E$ equals the interior angle at the opposite vertex $C$: $\angle AED = \angle DCB = 40^{\circ}$. Then $\triangle ADE$ has $\angle A = 70^{\circ}$ and $\angle AED = 40^{\circ}$, giving $\angle ADE = 70^{\circ}$. The exterior angle of $\triangle ADE$ at $D$ along $\overline{BD}$ is then $110^{\circ}$, and that exterior angle equals $\angle BFC$ via the vertical-angle / linear-pair chain — confirming $(D)$.
CCSS standards used (min grade 8)
4.G.A.1Draw points, lines, line segments, rays, angles, and identify in figures (Sketching $\triangle ABC$, the circle on diameter $\overline{BC}$, points $D, E$, and the diagonals meeting at $F$.)4.MD.C.6Measure angles in whole-number degrees using a protractor (Matching $110^{\circ}$ to the multiple-choice value.)7.G.B.5Use facts about supplementary, complementary, vertical, and adjacent angles (Computing $\angle BCE = 20^{\circ}$ and $\angle DBC = 50^{\circ}$ inside the two Thales right triangles.)8.G.A.5Use informal arguments to establish facts about angle sum and exterior angles (Splitting $\triangle ABC$ into base angles of $70^{\circ}$ and concluding $\angle BFC = 110^{\circ}$ in $\triangle BFC$.)
⭐ This AMC 10 problem only needs Grade 8 angle-chasing you already know! Because $BC$ is a diameter, the angles $\angle BEC$ and $\angle BDC$ are right angles (Thales). The isosceles triangle gives base angles $70^{\circ}$. Inside the two right triangles: $\angle BCE = 20^{\circ}$ and $\angle DBC = 50^{\circ}$. Then $\triangle BFC$ closes: $180^{\circ} - 50^{\circ} - 20^{\circ} = \mathbf{110^{\circ}}$, answer $(D)$.
⭐ This AMC 10 problem only needs Grade 8 angle-chasing you already know! Because $BC$ is a diameter, the angles $\angle BEC$ and $\angle BDC$ are right angles (Thales). The isosceles triangle gives base angles $70^{\circ}$. Inside the two right triangles: $\angle BCE = 20^{\circ}$ and $\angle DBC = 50^{\circ}$. Then $\triangle BFC$ closes: $180^{\circ} - 50^{\circ} - 20^{\circ} = \mathbf{110^{\circ}}$, answer $(D)$.
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