AMC 10 · 2019 · #13

Grade 8 geometry-2d

Archetype Arithmetic Expression Decomposition

inscribed-angleisosceles-triangleangle-sum-trianglearc-measure identify-subproblemscasework ↑ Prerequisites: inscribed-angleisosceles-triangleangle-sum-triangle

📏 Medium solution 💡 3 insights

Problem

Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$ . Construct the circle with diameter $\overline{BC}$ , and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB}$ , respectively. Let $F$ be the intersection of the diagonals of the quadrilateral $BCDE$ . What is the degree measure of $\angle BFC ?$

$\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120$

Pick an answer.

(A)

(B)

100

(C)

105

(D)

110

(E)

120

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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

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

#1 Draw a Diagram 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 $F$.

$$\text{circle on diameter } \overline{BC}; \;\; D \in \overline{AC},\; E \in \overline{AB}$$

💡 Grade 4 figures: a clean labeled picture makes every angle visible.

#7 Identify Subproblems 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 $\angle BAC = \angle ABC = 70^{\circ}$.

$$\angle BAC = \angle ABC = \tfrac{180^{\circ} - 40^{\circ}}{2} = 70^{\circ}$$

💡 Grade 8 angle sums: the two equal base angles split the leftover $140^{\circ}$ evenly.

#7 Identify Subproblems 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 $\angle BCE = 180^{\circ} - 90^{\circ} - 70^{\circ} = 20^{\circ}$.

$$\angle BEC = 90^{\circ},\;\; \angle BCE = 180^{\circ} - 90^{\circ} - 70^{\circ} = 20^{\circ}$$

💡 Grade 7 angle facts: a right angle and a known angle force the third in a triangle.

#7 Identify Subproblems 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 $\angle DBC = 180^{\circ} - 90^{\circ} - 40^{\circ} = 50^{\circ}$.

$$\angle BDC = 90^{\circ},\;\; \angle DBC = 180^{\circ} - 90^{\circ} - 40^{\circ} = 50^{\circ}$$

💡 Grade 7 angle facts: another right angle peels off another tiny triangle.

#7 Identify Subproblems 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 $180^{\circ}$: $\angle BFC = 180^{\circ} - 50^{\circ} - 20^{\circ} = 110^{\circ}$.

$$\angle BFC = 180^{\circ} - 50^{\circ} - 20^{\circ} = 110^{\circ}$$

💡 Grade 8: the diagonal-intersection triangle is just one more angle-sum.

#3 Eliminate Possibilities 4.MD.C.6 Step 6

Match $110^{\circ}$ to the choices: $(D)$.

$$110^{\circ} \;\Rightarrow\; \textbf{(D)}$$

💡 Grade 4 angle measure: pick the matching degree value.

[1] #1 4.G.A.1 Sketch the figure. Draw $\triangle ABC$ with $A$ at the top and $BC$ at the bott

[2] #7 8.G.A.5 Find the base angles of $\triangle ABC$. The triangle is isosceles with $BC = AC

[3] #7 7.G.B.5 Use Thales' theorem at $E$. Because $\overline{BC}$ is the diameter and $E$ is o

[4] #7 7.G.B.5 Use Thales at $D$. Because $D$ is on the circle and $\overline{BC}$ is the diame

[5] #7 8.G.A.5 Now look at $\triangle BFC$. The angle at $B$ is $\angle FBC$, which is the same

[6] #3 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.1 Draw 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.6 Measure angles in whole-number degrees using a protractor (Matching $110^{\circ}$ to the multiple-choice value.)
7.G.B.5 Use 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.5 Use 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 $(D)$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 8)

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