AMC 10 · 2011 · #17
Grade 8 geometry-2dIn the given circle, the diameter EB is parallel to DC, and AB is parallel to ED. The angles AEB and ABE are in the ratio 4 : 5. What is the degree measure of angle BCD?
Pick an answer.
AMC 10 2011 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 a circle, $\overline{EB}$ is a diameter and it is parallel to chord $\overline{DC}$. Chord $\overline{AB}$ is parallel to chord $\overline{ED}$. The two angles of triangle $AEB$ at $E$ and $B$ are in the ratio $\angle AEB : \angle ABE = 4 : 5$. Find the degree measure of $\angle BCD$.
Givens: $A, B, C, D, E$ all lie on one circle; going around, the order is $B, C, D, E, A$; $\overline{EB}$ is a diameter of the circle; $\overline{EB} \parallel \overline{DC}$; $\overline{AB} \parallel \overline{ED}$; $\angle AEB : \angle ABE = 4 : 5$; Choices: (A) $120$, (B) $125$, (C) $130$, (D) $135$, (E) $140$
Unknowns: The degree measure of $\angle BCD$
Understand
Restated: In a circle, $\overline{EB}$ is a diameter and it is parallel to chord $\overline{DC}$. Chord $\overline{AB}$ is parallel to chord $\overline{ED}$. The two angles of triangle $AEB$ at $E$ and $B$ are in the ratio $\angle AEB : \angle ABE = 4 : 5$. Find the degree measure of $\angle BCD$.
Givens: $A, B, C, D, E$ all lie on one circle; going around, the order is $B, C, D, E, A$; $\overline{EB}$ is a diameter of the circle; $\overline{EB} \parallel \overline{DC}$; $\overline{AB} \parallel \overline{ED}$; $\angle AEB : \angle ABE = 4 : 5$; Choices: (A) $120$, (B) $125$, (C) $130$, (D) $135$, (E) $140$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable, #7 Identify Subproblems, #3 Eliminate Possibilities
This is an angle-chase on a labeled circle, so Tool #1 (work the diagram) is primary: mark every angle the givens force, then follow them to $\angle BCD$. Tool #4 names the ratio parts as $4x$ and $5x$ so the $4:5$ split becomes one equation. Tool #7 breaks the chase into clean stages — first the right angle from the diameter, then the parallel-line angle, then the cyclic-quadrilateral angle. Tool #3 matches the final number to a choice.
Execute — Answer: C
8.G.A.5 Step 1 Diameter forces a right angle
- Point $A$ sits on the circle and looks across the diameter $\overline{EB}$.
- Any point on a circle that looks across a diameter sees it as a right angle, so $\angle EAB = 90^\circ$.
- That makes $\triangle AEB$ a right triangle with the right angle at $A$.
💡 A corner that spans a full diameter is always a square corner.
8.G.A.5 Step 2 The other two angles add to 90
- The three angles of $\triangle AEB$ add to $180^\circ$.
- Since one of them is $90^\circ$, the other two — $\angle AEB$ and $\angle ABE$ — must add to $180^\circ - 90^\circ = 90^\circ$.
💡 Take the right angle out of $180^\circ$ and $90^\circ$ is left to share.
7.EE.B.4 Step 3 Split 90 in the ratio 4 : 5
- Write the two angles as $\angle AEB = 4x$ and $\angle ABE = 5x$, matching the $4:5$ ratio.
- Their sum is $90^\circ$, so $4x + 5x = 90^\circ$, giving $9x = 90^\circ$ and $x = 10^\circ$.
- Then $\angle AEB = 40^\circ$ and $\angle ABE = 50^\circ$.
💡 Ratio $4:5$ is $9$ equal shares of the $90^\circ$, so each share is $10^\circ$.
8.G.A.5 Step 4 Slide the angle along parallel chords
- The chords $\overline{AB}$ and $\overline{ED}$ are parallel, and the diameter $\overline{EB}$ crosses both of them like a transversal.
- So $\angle ABE$ (at $B$, between $\overline{BA}$ and $\overline{BE}$) and $\angle DEB$ (at $E$, between $\overline{ED}$ and $\overline{EB}$) are alternate interior angles, which are equal.
- Therefore $\angle DEB = \angle ABE = 50^\circ$.
💡 Parallel chords cut by the same line make matching Z-angles.
7.G.B.5 Step 5 Opposite angles of the inscribed quadrilateral
- The four points $B, C, D, E$ lie on the circle and form quadrilateral $BCDE$.
- In any quadrilateral inscribed in a circle, opposite angles add to $180^\circ$.
- The angle at $C$ (which is $\angle BCD$) and the angle at $E$ (which is $\angle DEB$) are opposite, so $\angle BCD + \angle DEB = 180^\circ$.
- Hence $\angle BCD = 180^\circ - 50^\circ = 130^\circ$.
💡 In a circle-inscribed quadrilateral, each pair of far corners totals a straight angle.
4.NBT.A.2 Step 6 Match to a choice
$\angle BCD = 130^\circ$ matches choice (C).
💡 Read off the answer that equals $130$.
8.G.A.5 Point $A$ sits on the circle and looks across the diameter $\overline{EB}$. Any 8.G.A.5 The three angles of $\triangle AEB$ add to $180^\circ$. Since one of them is $90 7.EE.B.4 Write the two angles as $\angle AEB = 4x$ and $\angle ABE = 5x$, matching the $4 8.G.A.5 The chords $\overline{AB}$ and $\overline{ED}$ are parallel, and the diameter $\ 7.G.B.5 The four points $B, C, D, E$ lie on the circle and form quadrilateral $BCDE$. In 4.NBT.A.2 $\angle BCD = 130^\circ$ matches choice (C). Review
Reasonableness: The answer should be an obtuse angle: $C$ sits on the short arc near the top while $B$ and $D$ are spread wide, so $\angle BCD$ looking back across them is clearly more than $90^\circ$ — $130^\circ$ fits. Cross-check with arcs: the inscribed angle $\angle BCD$ cuts off the far arc $BD$ (through $E$ and $A$). Since $\angle DEB = 50^\circ$ is an inscribed angle on arc $BD$ (the near side), the far arc measures $360^\circ - 2(50^\circ) = 260^\circ$, and half of that is $130^\circ$ — the same value, so the answer is consistent. The given $\overline{EB} \parallel \overline{DC}$ pins down exactly where $C$ lands, but the opposite-angle rule fixes $\angle BCD$ for any $C$ on that arc, which is why we did not need it directly.
Alternative: Tool #4 (Introduce a Variable) with arcs: let the circle's $360^\circ$ of arc be split by the five points and write each inscribed angle as half its subtended arc. From $\angle ABE = 50^\circ$ you get arc $AE = 100^\circ$; the two parallel-chord conditions force arc $BD = 100^\circ$ as well, so the arc $BAED$ opposite $C$ is $360^\circ - 100^\circ = 260^\circ$, giving $\angle BCD = \tfrac{1}{2}(260^\circ) = 130^\circ$.
CCSS standards used (min grade 8)
8.G.A.5Use informal arguments to establish facts about angle sum and exterior angles (The right angle across the diameter, the triangle angle sum giving $\angle AEB + \angle ABE = 90^\circ$, and the alternate-interior-angle step $\angle DEB = \angle ABE$.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Writing the two angles as $4x$ and $5x$ and solving $9x = 90^\circ$ for $x = 10^\circ$.)7.G.B.5Use facts about supplementary, complementary, vertical, and adjacent angles (Using opposite angles of the inscribed quadrilateral $BCDE$ summing to $180^\circ$ to get $\angle BCD = 180^\circ - 50^\circ$.)4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Matching the computed value $130$ to answer choice (C).)
⭐ A corner that spans a diameter is a right angle, a ratio just splits an angle into equal shares, and opposite corners of a circle-inscribed quadrilateral add to $180^\circ$ — chain those and $\angle BCD = 130^\circ$.
⭐ A corner that spans a diameter is a right angle, a ratio just splits an angle into equal shares, and opposite corners of a circle-inscribed quadrilateral add to $180^\circ$ — chain those and $\angle BCD = 130^\circ$.
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